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iS^^°- \>/ ^^fe' %/ .-^T: *--*^ -'Jfe: V 







THE MANUFACTURE 

OF 

PULP AND PAPER 



VOLUME I 



Pulp and Paper Manufacture 

IN FIVE VOLUMES 

An Official Work Prepared 
under the direction of the 

Joint Executive Committee of the 

Vocational Education Committees of the 

Pulp and Paper Industry of the 

United States and Canada 



Vol. I — Mathematics, How to Read 
Drawings, Physics. 
II — Mechanics and Hydraulics, 

Electricity, Chemistry. 
Ill — Preparation of Pulp. 
IV, V — Manufacture of Paper. 



THE 

MANUFACTURE 

OF 

PULP AND PAPER 

A TEXTBOOK OF MODERN PULP 
AND PAPER MILL PRACTICE 

Prepared Under the Direction of the Joint Executive 

Committee on Vocational Education Representing 

the Pulp and Paper Industry of the 

United States and Canada 




VOLUME I 

Arithmetic, Elementary Applied Mathematics, 
How TO Read Drawings, Elements of Physics 

BY 

J. J. CLARK, M.E. 

5 



First Edition 
Second Impression 



McGRAW-HILL BOOK COMPANY, Inc. 

NEW YORK: 370 SEVENTH AVENUE 

LONDON: 6 & 8 BOUVERIE ST., E. C. 4 

1921 



\\'>^ 



^6 



coptbight, 1921, bt the 

Joint Executive Committee of the Vocational Edtjcation Committees 

OF THE Pulp and Paper Industry. 



All Rights Reserved, 
Including Those of Translation. 






THE M: A P L E PKESS X O K It FA. 



PREFACE 



In numerous communities where night schools and extension 
classes have been started or planned, or where men wished to 
study privately, there has been difficulty in finding suitable 
textbooks. No books were available in English, which brought 
together the fundamental subjects of mathematics and element- 
ary science and the principles and practice of pulp and paper 
manufacture. Books that treated of the processes employed 
in this industry were too technical, too general, out of date, or so 
descriptive of European machinery and practice as to be unsuit- 
able for use on this Continent. Furthermore, a textbook was 
required that would supply the need of the man who must study 
at home because he could not or would not attend classes. 

Successful men are constantly studying; and it is only by 
studying that they continue to be successful. There are many 
men, from acid maker and reel-boy to superintendent and mana- 
ger, who want to learn more about the industry that gives them a 
livehhood and by study to fit themselves for promotion and in- 
creased earning power. Pulp and paper makers want to under- 
stand the work they are doing- — -the how and why of all the 
various processes. Most operations in this industry are, to some 
degree, technical, being essentially either mechanical or chemical. 
It is necessary, therefore, that the person who aspires to under- 
stand these processes should obtain a knowledge of the under- 
lying laws of Nature through the study of the elementary sciences 
and mathematics, and be trained to reason clearly and logically. 

After considerable study of the situation by the Committee 
on Education for the Technical Section of the Canadian Pulp 
and Paper Association and the Committee on Vocational Educa- 
tion for the Technical Association of the (U. S.) Pulp and Paper 
Industry, a joint meeting of these committees was held in Buffalo 



vi PREFACE 

in September, 1918, and a Joint Executive Committee was ap- 
pointed to proceed with plans for the preparation of the text, its 
pubhcation, and the distribution of the books. The scope of the 
work was defined at this meeting, when it was decided to provide 
for preHminary instruction in fundamental Mathematics and 
Elementary Science, as well as in the manufacturing operations 
involved in modern pulp and paper mill practice. 

The Joint Educational Committee then chose an Editor, 
Associate Editor, and Editorial Advisor, and directed the Editor 
to organize a staff of authors consisting of the best available men 
in their special lines, each to contribute a section dealing with his 
specialty. A general outline, with an estimated budget, was 
presented at the annual meetings in January and February, 1919, 
of the Canadian Pulp and Paper Association, the Technical 
Association of the Pulp and Paper Industry and the American 
Paper and Pulp Association. It received the unanimous approval 
and hearty support of all, and the budget asked was raised by 
an appropriation of the Canadian Pulp and Paper Association 
and contributions from paper and pulp manufacturers and allied 
industries in the United States, through the efforts of the 
Technical Association of the Pulp and Paper Industry. 

To prepare and publish such a work is a large undertaking; 
its successful accomplishment is unique, as evidenced by these 
volumes, in that it represents the cooperative effort of the Pulp 
and Paper Industry of a whole Continent. 

The work is conveniently divided into sections and bound into 
volumes for reference purposes; it is also available in pamphlet 
form for the benefit of students who wish to master one part 
at a time, and for convenience in the class room. This latter 
arrangement makes it very easy to select special courses of 
study; for instance, the man who is specially interested, say, in 
the manufacture of pulp or in the coloring of paper or in any 
other special feature of the industry, can select and study the 
special pamphlets bearing on those subjects and need not study 
others not relating particularly to the subject in which he is 
interested, unless he so desires. The scope of the work enables 
the man with but little education to study in the most efficient 
manner the preliminary subjects that are necessary to a 
thorough understanding of the principles involved in the manu- 
facturing processes and operations; these subjects also afford an 
exce)lent review and reference textbook to others. The work 



PREFACE vii 

is thus especially adapted to the class room, to home study, 
and for use as a reference book. 

It is expected that universities and other educational agencies 
will institute correspondence and class room instruction in 
Pulp and Paper Technology and Practice with the aid of these 
volumes. The aim of the Committee is to bring an adequate 
opportunity for education in his vocation within the reach of 
every one in the industry. To have a vocational education 
means to be familiar with the past accomplishments of one's trade 
and to be able to pass on present experience for the benefit of 
those who will follow. 

To obtain the best results, the text must be diligently studied; 
a few hours of earnest application each week will be well repaid 
through increased earning power and added interest in the daily 
work of the mill. To understand a process fully, as in making 
acid or sizing paper, is like having a light turned on when one 
has been working in the dark. As a help to the student, many 
practical examples for practice and study and review questions 
have been incorporated in the text ; these should be conscientiously 
answered. 

The Editor extends his sincere thanks to the Committee and 
others, who have been a constant support and a source of in- 
spiration and encouragement; he desires especially to mention 
Mr. George Carruthers, Chairman, and Mr. R. S. Kellogg, 
Secretary, of the Joint Executive Committee; Mr. J. J. Clark, 
Associate Editor, Mr, T. J. Foster, Editorial Advisor, and Mr. 
John Erhardt of the McGraw-Hill Book Company, Inc. 

The Committee and the Editor have been generously assisted 
on every hand ; busy men have written and reviewed manuscript, 
and equipment firms have contributed drawings of great value 
and have freely given helpful service and advice. Among these 
kind and generous friends of the enterprise are: Mr. 0. Bache- 
Wiig, Mr. James Beveridge, Mr. J. Brooks Beveridge, Mr. H. 
P. Carruth, Mr. Martin L. Grifiin, Mr. H. R. Harrigan, Mr. 
Arthur Burgess Larcher, Mr. J. 0. Mason, Mr. Elis Olsson, Mr. 
George K. Spence, Mr. Edwin Sutermeister, Mr. F. G. Wheeler, 
and American Writing Paper Co., Dominion Engineering Works, 
E. I. Dupont de Nemours Co., F. C. Huyck & Sons, Hydraulic 
Machinery Co., Improved Paper Machinery Co., E. D. Jones 
& Sons Co., A. D. Little, Inc., National Aniline and Chemical 
Works, Process Engineers, Pusey & Jones Co., Rice, Barton & 



viii PREFACE 

Fales Machine and Iron Works, Ticonderoga Paper Co., Waterous 
Engine Works Co., and many others. 

J. Newell Stephenson, 

Editor 

For the 

Joint ExECUTrvE Committee on Vocational Education, 

George Carruthers, Chairman, R. S. Kellogg, Secretary, 

T. L. Crossley, G. E. Williamson, C. P. Winslow. 

Representing the Technical Sec- Representing the Technical As- 
tion of the Canadian Pulp and Paper sociation of the (U. S.) Pulp and 
Association Paper Industry. 

T. L. Crossley, Chairman, George E. Williamson, Chairman, 

George Carruthers, Hugh P. Baker, 

A. P. CosTiGANE, Henry J. Guild, 

Dan Daverin, R. S. Kellogg, 

C. Nelson Gain, Otto Kress, 

J. N. Stephenson. W. S. Lucey, 

C. P. Winslow. 



CONTENTS 



Paqh 

. Preface v 

SECTION 1 
Arithmetic 

PART I 

Notation and Enumeration 1-10 

Addition and Subtraction . 11-21 

Multiplication and Division. . , 22-34 

Some Properties op Numbers 34—39 

Examination Questions 41-42 

PART II 

Greatest Common Divisor and Least Common Multiple . . . 43-47 

Fractions 48-64 

Involution 64-67 

Decimals and Decimal Fractions 67-76 

Signs of Aggregation 76-77 

Ratio and Proportion 77-85 

Examination Questions 87-88 

PART III 

Square Root 89-98 

Percentage 98-106 

Compound Numbers 106-113 

The Metric System 113-124 

The Arithmetical Mean 125-126 

Examination Questions 127-128 

SECTION 2 
Elementary Applied Mathematics 

PART I 

Mathematical Formulas 1-4 

Algebraic Addition, Subtraction, Multiplication, and 

Division 4-18 

Equations 18-27 

Accuracy in Calculation 27-33 

Approximate Method for Finding Roots 33-39 

Examination Questions 41-42 

ix 



X CONTENTS 

PART II 

Page 

Mensueation of Plane Figtjres 43-47 

Triangles 50-60 

Quadrilaterals 61-66 

Regular Polygons 66-68 

The Circle 68-89 

Inscribed and Circumscribed Polygons 89-96 

Examination Questions 97-100 

PART III 

Mensuration of Solids 101-130 

Similar Figures 131-137 

Symmetrical Figures 137-143 

Examination Questions 145-146 

SECTION 3 
How to Read Drawings 

Representing Solids ON Planes 1-11 

Special Features Pertaining to Drawings 12-23 

Reading Drawings — Visualizing the Object 23-28 

Some Examples in Reading Drawing 28 



SECTION 4 
Elements of Physics 

PART I 

Matter and Its Properties 1-6 

Motion and Velocity 6-9 

Force, Mass and Weight 9-14 

Work and Energy 14r-17 

Hydrostatics — Pascal's Law 18-30 

Buoyancy and Specific Gravity 31-39 

Capillarity 39-41 

Pneumatics 42-54 

Examination Questions 55-56 

PART II 

Properties op Perfect Gases 57-66 

Nature op Measurement of Heat 66-80 

Change op State 81-90 

Light 90-117 

Examination Questions 119-120 

Index 121-132 



SECTION 1 

AEITHMETIC 

PART 1 



NOTATION AND ENUMERATION 



DEFINITIONS 

1. A unit is the standard by which anything is measured. For 
example if it were desired to measure the length of a room, it 
could be done with a tape line, a rule, a stick of some convenient 
length, or anything else that would form a basis of comparison. 
Suppose, for example, a broom- 
stick were used ; the broomstick 
would be laid on the floor, with 
one end a against the wall at one 
side of the room and with the 
other end h extending towards 
the opposite side of the room, 
as shown in Fig. 1. The point h 
would be marked and the broom- 
stick would be shifted so as to 
occupy the position he, then to 
the position cd, and so on until 
the other wall was reached. 
Each of the lengths ah, he, etc. 

is equal in length to the broomstick, which is in this case the unit 
of length. The word that denotes how many times the room is 
longer than the broomstick is numher. In other words number 
means an aggregation, or collection, of units; a number may 
also mean a single unit or a part of a unit — it refers in all cases to 
the complete measurement. 

As another example, consider the compensation that a man 
receives for his work. For working a certain number of hours he 
receives a certain number of dollars; here two units are involved, 
§1 1 



a 


ft 


c 


d 


e 


/ 


1 


s 


> 


\ t 


! 





Fig. 1. 



2 ARITHMETIC §1 

hours and dollars. The unit of hours is established or arrived at 
by dividing or splitting up a day (another unit) into twenty-four 
equal parts and calling one of these parts one hour. The unit of 
money is established by law; it is called in the United States of 
America one dollar, and is equivalent in value to about twenty- 
three and one-fourth grains (another unit) of pure gold. Canada 
has a unit of the same name and of practically the same value. 

2. Three different kinds of units, according to their origin, 
have been mentioned, the broomstick, which may be called an 
expediency unit; the hour, which is an international unit for 
measuring time; and the dollar, which is a legal unit, or one estab- 
lished by law. According to their use, units have many names, 
and anything that can be expressed as a certain number of units 
is called a quantity. The science that treats of quantity and its 
measurement is called mathematics. 

3. The difference between quantity and number is this : quantity 
is a general term and is applied to anything that can be measured 
or expressed in units; number is simply a term applied to a unit 
or a collection of units and is usually restricted to expressions 
containing only figures. 

4. The act or process by which numbers are used to reckon, 
count, estimate, etc. is called computation or calculation. 

5. Arithmetic is that branch of mathematics which treats of 
numbers and their use in computation. 

Numbers are divided into two general classes, according to their 
signification: abstract numbers and concrete numbers. 

6. An abstract number is one not applied to any object or 
quantity, as three, twelve, one hundred seventeen, etc. 

7. A concrete number is one relating to a particular kind of 
object or quantity, as five dollars, eleven pounds, seven hours, etc. 

According to their units, numbers are also divided into two 
general classes: 

8. Like numbers are numbers having the same units: for ex- 
ample, seven hours, twelve hours, and three hours are like numbers, 
since they all have the same unit — one hour; fourteen, seventeen 
and twenty are also like numbers, the unit of each being one. All 
abstract numbers are like numbers. 

9. Unlike numbers are numbers having different units; for 
example, five dollars, eight hours, and ten pounds are unlike 



§1 NOTATION AND ENUMERATION 3 

numbers since their units are unlike, being respectively one dollar, 
one hour, and one pound. 



NOTATION 

In order that numbers may be recorded, some method of writing 
and reading them must be devised. 

10. Notation is the word used to express the act of and the 
result obtained by writing numbers. 

11. Numeration is the act or process of reading numbers that 
have been written. 

12. Notation is accomplished in three ways: (1) by words; (2) 
by letters; (3) by figures. The first method, by words, is never 
used in computation, only the second and third methods having 
ever been employed for this purpose. The second method, called 
the Roman notation, is very seldom used at the present time, its 
only use is for numbering, as the indexes, chapters, etc. of books, 
and in a few cases for dates. The third method, the Arabic 
notation, employs certain characters, called figures, and is in 
universal use today. 



THE FIRST OR WORD METHOD 

13. This is really only the names of the different numbers. 
The first twelve numbers have distinct names; beyond these all 
are formed according to a definite system that is readily learned. 
The names of the first twelve numbers are: one, two, three, 
four, five, six, seven, eight, nine, ten, eleven, twelve. The next 
seven numbers all end in teen, which means and ten, the first part of 
the word being derived from the words three, four, etc. up to nine; 
they are thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, 
and nineteen. Thirteen means three and ten, fourteen means 
four and ten, etc. The name of the next number is twenty, and 
the names of the next nine numbers are all compound words, 
having for their first element the word twenty and for their second 
element the words one, two, etc. to nine. Thus, twenty-one, 
twenty-two, etc. to twenty-nine. The word twenty means two 
tens. The next number is thirty, which means three tens, and the 
next nine are thirty-one, thirty-two, etc. to thirty-nine. Then 
follow forty, fifty, sixty, seventy, eighty, and ninety, with their 



4 ARITHMETIC §1 

combinations of one, two, etc. to nine. The number following 
ninety-nine is one hundred. The numbers are then repeated, 
with the element one hundred added, one-hundred-one, one- 
hundred-two, etc., one-hundred-twenty-one, one-hundred-twenty- 
two, etc. to one-hundred-ninety-nine, the next number being 
two-hundred. The notation is continued in this manner to nine- 
hundred-ninety-nine, the next number being one-thousand; then 
follow one-thousand-one, one-thousand-two, etc., one-thousand- 
one-hundred-one, one-thousand-one-hundred-two, etc. to nine- 
thousand-nine-hundred-ninety-nine, the next number being ten- 
thousand. Then follow ten-thousand-one, etc. to nine-hundred- 
thousand-nine-hundred-ninety-nine, the next number being called 
one-million. The notation is then carried on one-million-one, etc. 
up to one-thousand-million, which in the United States, France, 
and the majority of other countries is called a billion; a thousand 
billion is a trillion, etc. In Great Britain and a few other coun- 
tries, a billion is a million-million, a trillion a million-billion, etc. 



THE ROMAN NOTATION 

14. The Roman notation uses the letters I, V, X, L, C, D, and 
M. The values of the letters when standing alone are : I is one, 
V is five, X is ten, L is fifty, C is one-hundred, D is five-hundred, 
M is one-thousand. Formerly other letters were used, but these 
are all that are employed now. Numbers are expressed by letters 
in accordance with the following principles : 

I. Repeating a letter repeats its value. 

Thus, I is one, II is two. III is three, XX is two tens or twenty, 
XXX is three tens or thirty, CC is two hundred, MM is two- 
thousand, etc. 

V, D, and L are never repeated; only I, X, C, and M are ever 
used more than once in any number, except when they precede a 
letter of higher value. 

II. If a letter precedes one of greater value, their difference is 
denoted; if it follows a letter of greater value, their sum is denoted. 

Thus, IV is four, VI is six, IX is nine, XII is twelve, XL is forty, 
LX is sixty, XC is ninety, etc. Only one letter of lower value is 
allowed to precede one of higher value, and in general, the only 
letters used are I and X. Further, I precedes only V and X, and 
X precedes only L and C. Occasionally, however, C is used in 



§1 NOTATION AND ENUMERATION 5 

this manner before M ; as for example, in the date nineteen-hun- 
dred-four, when instead of writing MDCCCCIV the shorter form 
MCMIV is used. Pope Leo XIII wrote the date eighteen- 
hundred ninety-five MDCCCVC instead of MDCCCXCV. On 
the Columbus memorial in front of the Union Station, Washington, 
D. C, the year of his birth is expressed by MCDXXXVI, which, 
of course, is read fourteen-hundred-thirty-six. 

III. A bar placed over a letter multiples its value by one-thousand. 

Thus, X is ten-thousand, L is fifty-thousand, XCDXVII is 
ninety-thousand five-hundred seventeen. In the last expression, 
it might be thought that D was preceded by X and C; this is not 
the case, however, as XC is treated as a single letter having a 
value of ninety-thousand. 

The following table illustrates the foregoing principles and 
applications of the Roman notation : 

VIII is eight 

XII is twelve 

XVIII is eighteen 

XXIX is twenty-nine 

XXXV is thirty-five 

XLIV is forty-four 

LXXVI is seventy-six 

XCII is ninety-two 

CXLIV is one-hundred forty-four 

CCCCL is four-hundred fifty 

IXLX is nine-thousand sixty 

C is one-hundred-thousand 

D is five-hundred-thousand 

M is one-million 



THE ARABIC NOTATION 

15. The Arabic notation employs ten characters, called 
figures, to represent numbers; these characters together with 
their names are : 

123456789 

naught one two three four five six seven eight nine 

cipher 
zero 

The first figure (0) is called naught, cipher, or zero, and has no 
value. The other nine figures are called digits, and each has 
the value denoted by its name. 



6 ARITHMETIC §1 

16. For numbers higher than nine it is necessary to use two or 
more figures to express them. Thus ten is expressed by the com- 
bination 10, the cipher indicating that there are no units in the 
right-hand figure; 11 is eleven, 12 is twelve, 13 is thirteen, etc. up 
to 19 or nineteen. All these expressions mean ten and one, ten 
and two, etc. Twenty is written 20, twenty-one, 21, etc.; thirty 
is written 30, thirty-three, 33, etc.; and so on up to 99 or ninety- 
nine. One-hundred is written 100, the right-hand cipher mean- 
ing that there are no units and the next that there are no tens. 
Five-hundred-forty-two is written 542, five-hundred-two is writ- 
ten 502, five-hundred-forty is written 540. In a similar manner 
one-thousand is written 1000, ten-thousand is written 10000, 
one-hundred-thousand is written 100000, one-million is written 
1000000, etc. 

17. As stated before the cipher has no value, but it plays a very 
important part in expressing numbers by figures ; it not only indi- 
cates the absence of a digit from the place it occupies, but it also 
locates the digits with respect to one another. 

Any number, as for example, 5642, is equivalent to 5000 and 
600 and 40 and 2; that is, in this case, it is equivalent to four sepa- 
rate numbers, three of which are single digits, all followed by one 
or more ciphers. Such a number might be written 5000 + 600 
+ 40 + 2, but this is not necessary, since the ciphers can be 
omitted without impairing the result, with the additional ad- 
vantage of saving space and figures. 

18. The number denoted by a figure in any whole number is 
determined by writing the figure and then writing after it as 
many ciphers as there are figures to the right of the one selected. 
Thus, in the number 2645893, the number denoted by the figure 
6 is 600000 or six-hundred-thousand ; the number denoted by the 
figure 5 is 5000 or five-thousand; etc. 



NUMERATION 

19. For convenience numbers are divided into orders and 
periods. A figure belongs to the first order when it occupies the 
units place; it belongs to the second order when it occupies the 
tens place; etc. In the number 2645893, 2 belongs to the seventh 
order, 6 to the sixth order, etc. Numbers are divided by commas 
into ^periods of three figures each, beginning with the first order, 



§1 NOTATION AND ENUMERATION 7 

to assist in reading them; the periods are named according to the 
name of the lowest order in the periods. Thus, the number 
506,273,985,114 contains four full periods; beginning at the right 
and going to the left, the names of the periods are, respectively, 
units, thousands, millions, billions. The names of periods and 
orders represented by the figures in the above number are 
conveniently shown in the table below. 



Fourth Period. 
Billions. 


Third Period. 
Millions. 


Second Period. 
Thousands. 


First Period. 
Units. 


C 

^ 

d 

i 

6 

i 


a 
.2 

W 
S 

u 

o 

g 
;> 


TO 

.2 

o 


m 

a 
_2 

s 

•xi 

c 

a; 

o 
c 


aj 

C 

s 

1 

el 

£ 

o 

■+3 

-a 


CO 

1 

O 

1 

CO 


m 
eS 

DO 

o 

1 

w 
o 


m 

d 

O 

d 

o 


02 

d 
o 

o 

d 


T3 

M 

CD 

o 


d 

s 

O 

1 


03 

1 

o 
2 



6, 



3, 



9 



5, 



1 



In pointing off these figures, begin at the right-hand figure and 
count — units, tens, hundreds; the next group of three figures is 
thousands; therefore, insert a comma (,) before beginning with 
them. Beginning at the figure 5, say thousands, ten thousands, 
hundred thousands, and insert another comma; next read millions, 
ten millions, hundred millions (insert another comma) ; lastly, read 
billions, ten billions, hundred billions. 

20. A number is read by beginning at the left and reading the 
figures composing each period as though it formed a number by 
itself and affixing the name of the period, taking each period in 
succession from left to right. Thus, the above number is read: 
five-hundred-six billion two-hundred seventy-three million nine- 
hundred-eighty-five thousand one-hundred-fourteen. Note that the 



8 ARITHMETIC §1 

name of the units period is not pronounced — it is always 
understood. 

Reading a line of figures in this manner is called numeration; 
and when the numeration is changed back to figures, it is called 
notation. 

For instance, the writing of the following figures, 

72,584,623 

would be the notation, and the numeration would be seventy-two 
million five-hundred-eighty-four thousand six-hundred-twenty-three. 

21. Note. — It is customary to leave the "s" off the words miUions, thou- 
sands, etc. in cases like the above, both in speaking and writing. 

22. Consider the number 500,000. If the 5, which is here a 
figure of the sixth order, be moved one place to the right, so as to 
occupy the fifth order, the number becomes 50,000, which is only 
one-tenth as large as before; while if the 5 be moved one place to 
the left, so as to occupy the seventh order, the number is ten 
times larger. In other words, moving a figure to the right de- 
creases the number it represents ten times for each order it 
passes into; and moving the figure to the left increases the number 
represented by the digit ten times for each order passed into. 
Thus, 5 is one-tenth of 50 and 50 is ten times 5; 50 is one-tenth of 
500 and 500 is ten times 50; etc. 

23. Integers and Decimals. Any figure, say 5, represents the 
number of units signified by its name, when standing alone. 
Moving it one place to the left, it represents five tens or fifty, and 
is written 50; moving it one place to the right, it represents five- 
tenths, and is written 0.5. Moving the figure two places to the 
left of its first position, it represents five hundred, written 500; if 
moved two places to the right, it represents five hundredths, 
written 0.05. Now 0.5 means that this number is only one-tenth 
as large as 5, and 0.05 means that this number is only one-tenth 
as large as 0.5 or one-hundredth as large as 5. This notation 
may be continued to the right as far as desired. 

24. The dot is called the decimal point ; it is used to point out 
the unit figure, and the part of the number to the right of the 
unit figure is called a decimal. All numbers not having any 
digits to the right of their unit figures (and which do not contain 
a common fraction) are called whole numbers or integers. The 
numbers dealt with heretofore have all been integers. The part 



§1 NOTATION AND ENUMERATION 9 

of a number to the left of the decimal point is called the integral 
part of the number. 

25. Decimals are read in much the same manner as integers. 
The only difference is that after the number has been read as 
though it were an integer, the name of the lowest (right-hand) 
order is pronounced with the letters ths added to the name. Thus, 
0.52976 is read fifty-two thousand nine-hundred-seventy-six 
hundred-thousandths; 0.529 is read five-hundred-twenty-nine 
thousandths. 

26. In using decimals be careful to mark the units place by 
placing the decimal point immediately to the right of the figure in the 
units place. Then continue the notation to the right of the units 
place precisely as to the left of it. The resemblance between the 
names of the places on the two sides of the units place is shown by 
the following table, in which the differences are marked by 
italics. 



-s 




















T) 























c 


O 


ai 


CO 










5 


to 


Co 

C 
03 
CO 

3 


O 
+1 


-o 


13 


T3 
d 
a 

m 

;=! 
o 


m 








-S 


-a 

m 
o 




(D 


O 


<U 








a> 


O 


<D 


^ 


jH 


u 






«0 


>H 


Xi 


t-> 


.0 


+3 
+3 


el 

13 


00 

03 
-fi 


'3 


-f3 




g 

+3 


d 

3 

rd 



27. In this (the Arabic) system of notation, each place or order 
is said to be higher than any place to the right of it, and lower 
than any place to the left of it. Thus, in the number 43.127, the 
highest place is tens, and the lowest place is thousandths. 

28. In writing and printing, decimals are frequently written 
.5, .529 etc. instead of 0.5, 0.529, etc. The latter way is the 
better and more correct, and the student is advised to write the 
cipher in all cases; for, if the decimal point should fail to print, be 
indistinct, or through carelessness not written, the cipher will 
frequently serve to distinguish between integers and decimals. 
It will be found, however, that the cipher has frequently been 
omitted throughout this textbook to accustom the reader to 
both methods of writing decimals. 

29. A number, part of which is an integer and part a decimal 
is called a mixed number. A mixed number is read by reading 



10 ARITHMETIC §1 

the integral part first, pronouncing the word and, and then 
reading the decimal part. Thus, the number given in Art. 26, 
or 543,210.12345 is read five-hundred-forty-three thousand 
two-hundred-ten and twelve-thousand-three-hundred-forty-five 
hundred-thousandths. 

The abbreviation "Art." means Article, and refers to the 
numbered paragraphs or sections throughout this textbook. The 
plural is written Arts.; thus, Arts. 1-10, means articles 1 to 10, 
both inclusive. 

Decimals are seldom pointed off into periods, especially in 
English-speaking countries. It may be done, however, (if 
desired) in the same manner as integers, beginning at the unit 
figure. Thus, 0.0000000217 would be pointed off 0.00,000,002,17. 
In reading this decimal, name the periods until the last is reached 
and then name the order of the right-hand figure; — in this case, 
units, thousandths, millionths, billionths, ten-billionths; — this 
determines the name of the lowest order. The number is then 
read at once as two-hundred-seventeen ten-billionths. 



EXAMPLES FOR PRACTICE 

Read the following numbers and write their names: 

(1) 20103 

(2) 1964727 

(3) 7926.4867 

(4) 0.0078314 

Write the following numbers : 

(5) Eight-hundred-sixty-six million forty-six-thousand seven-hundred- 
thirty-three. 

(6) Six-hundred-ninety-one and four-thousand-five-hundred-twenty-three 
ten-thousandths. 

(7) Seventeen and eighty-nine-thousand-two-hundred-seventy-six mill- 
ionths. 



30. The Four Fundamental Processes.— The processes of 
addition, subtraction, multiplication and division are called 
fundamental because all other processes employed in arithmetical 
computation are based on them; in other words, no computation 
can be made without employing one or more of these processes. 



§1 ADDITION AND SUBTRACTION 11 

ADDITION AND SUBTRACTION 



ADDITION 

31. Suppose it is desired to ascertain the length of a room, and 
that the distance from one end to the edge of a door nearest that 
end, the width of the door, and the distance from the other edge of 
the door to the other end of the room are known to be respect- 
ively, 6 feet, 3 feet, and 8 feet; then without getting a rule or 
other measure, the length of the room can be found by counting, 
starting with six, counting three more, getting nine, and then 
counting eight more after nine, getting seventeen. In other 
words, the combined lengths of 6 feet, 3 feet, and 8 feet is 17 feet, 
and the numbers 6, 3, and 8 taken together always give 17. The 
process of finding a single number equivalent in amount to two 
or more numbers taken together is called addition, and the single 
number found as the result of the process is called the sum. 
Thus, the sum of 6, 3, and 8 is 17. To add two or more numbers 
is to find their sum. 

32. The method just described for finding the sum of several 
numbers is impracticable except for very small numbers. For 
example, suppose it were desired to find the sum of $2,904, $976, 
and $3,520; not only would this take a very long time, but there 
would also be great danger of making a mistake. A method for 
finding the sum of a series of numbers will now be described. 

When it is desired to indicate that several numbers are to be 
added, they are written in a row and separated by Greek crosses. 
Thus, the fact that the numbers 6, 3, and 8 are to be added is 
indicated as follows: 6 + 3 + 8. The cross + is called the 
sign of addition or plus or plus sign, and the expression would be 
read, six plus three plus eight. If it is also desired to write the 
sum as indicating the result of the addition, this is done in the 
following manner. 

6 + 3 + 8 = 17 
The sign = is called the equality sign, it is read equals or is 
equal to, and signifies that the result of the operations indicated 
on one side is exactly equal to the number (or to the result of a 
series of operations indicated) on the other side. For example, 
5 + 2 + 10 = 17; hence, 6 + 3 + 8 = 5 + 2+ 10 = 17. This 
means that the result of the operation 6 + 3 + 8 is equal to the 
result of the operation 5 + 2 + 10 and that either sum is 17. 



12 



ARITHMETIC 



il 



An expression like 6 + 3 + 8= 17 is read, six plus three plus 
eight equals seventeen. 

33. Before one can add several numbers it is necessary that 
the sum of any two numbers from 1 to 9 be recognized as soon as 
seen. The following table gives the sum of any two numbers 
ADDITION TABLE 



1 and 1 is 2 


2 and 1 is 3 


3 and 1 is 4 


4 and 1 is 5 


1 and 2 is 3 


2 and 2 is 4 


3 and 2 is 5 


4 and 2 is 6 


1 and 3 is 4 


2 and 3 is 5 


3 and 3 is 6 


4 and 3 is 7 


1 and 4 is 5 


2 and 4 is 6 


3 and 4 is 7 


4 and 4 is 8 


1 and 5 is 6 


2 and 5 is 7 


3 and 5 is 8 


4 and 5 is 9 


1 and 6 is 7 


2 and 6 is 8 


3 and 6 is 9 


4 and 6 is 10 


1 and 7 is 8 


2 and 7 is 9 


3 and 7 is 10 


4 and 7 is 11 


1 and 8 is 9 


2 and Sis 10 


3 and 8 is 11 


4 and 8 is 12 


1 and 9 is 10 


2 and 9 is 11 


3 and 9 is 12 


4 and 9 is 13 


1 and 10 is 11 


2 and 10 is 12 


3 and 10 is 13 


4 and 10 is 14 


1 and 11 is 12 


2 and 11 is 13 


3 and 11 is 14 


4 and 11 is 15 


1 and 12 is 13 


2 and 12 is 14 


3 and 12 is 15 


4 and 12 is 16 


5 and 1 is 6 


6 and 1 is 7 


7 and 1 is 8 


8 and 1 is 9 


5 and 2 is 7 


6 and 2 is 8 


7 and 2 is 9 


8 and 2 is 10 


5 and 3 is 8 


6 and 3 is 9 


7 and 3 is 10 


8 and 3 is j 1 


5 and 4 is 9 


6 and 4 is 10 


7 and 4 is 11 


8 and 4 is 12 


5 and 5 is 10 


6 and 5 is 11 


7 and 5 is 12 


8 and 5 is 13 


5 and 6 is 11 


6 and 6 is 12 


7 and 6 is 13 


8 and 6 is 14 


5 and 7 is 12 


6 and 7 is 13 


7 and 7 is 14 


8 and 7 is 15 


5 and 8 is 13 


6 and 8 is 14 


7 and 8 is 15 


8 and 8 is 16 


5 and 9 is 14 


6 and 9 is 15 


7 and 9 is 16 


8 and 9 is 17 


6 and 10 is 15 


6 and 10 is 16 


7 and 10 is 17 


8 and 10 is 18 


5 and 11 is 16 


6 and 11 is 17 


7 and 11 is 18 


8 and 11 is 19 


5 and 12 is 17 


6 and 12 is 18 


7 and 12 is 19 


8 and 12 is 20 


9 and 1 is 10 


10 and 1 is 11 


Hand lis 12 


12 and 1 is 13 


9 and 2 is 11 


10 and 2 is 12 


11 and 2 is 13 


12 and 2 is 14 


9 and 3 is 12 


10 and 3 is 13 


11 and 3 is 14 


12 and 3 is 15 


9 and 4 is 13 


10 and 4 is 14 


11 and 4 is 15 


12 and 4 is 16 


9 and 6 is 14 


10 and 5 is 15 


11 and 5 is 16 


12 and 5 is 17 


9 and 6 is 15 


10 and 6 is 16 


11 and 6 is 17 


12 and 6 is 18 


9 and 7 is 16 


10 and 7 is 17 


11 and 7 is 18 


12 and 7 is 19 


9 and 8 is 17 


10 and 8 is 18 


11 and 8 is 19 


12 and 8 is 20 


9 and 9 is 18 


10 and 9 is 19 


11 and 9 is 20 


12 and 9 is 21 


9 and 10 is 19 


10 and 10 is 20 


11 and 10 is 21 


12 and 10 is 22 


9 and 11 is 20 


10 and 11 is 21 


11 and 11 is 22 


12 and 11 is 23 


9 and 12 is 21 


10 and 12 is 22 


11 and 12 is 23 


12 and 12 is 24 



Since has no value, the sum of any number and is the number itself; 
thus, 19 and is 19. 



§1 ADDITION AND SUBTRACTION 13 

from 1 to 12; it should be thoroughly committed to memory, and 
to do this the reader should propose to himself all kinds of com- 
binations of two numbers until as soon as the numbers are named 
he can give the sum almost unconsciously. When at work, when 
walking, or at any time when he has leisure he should practice by 
saying 8 and 9 is 17, 9 and 5 is 14, 5 and 11 is 16, 9 and 8 is 17, 
etc. until he can name the sum instantly and correctly. 

34. To Add a Digit and a Number Greater than 12. — The sum 
of a single digit and a number greater than 12 can always be 
obtained mentally. For example, consider the numbers 58 and 
9; write these numbers, one above the other, so that the unit 
figures stand in the same column, and draw a line under them. 

58 or 9 

1- 58 

Now add the unit figures, the sum being 17, and write the sum 
underneath the fine with the unit figure 7 standing in the same 
column as the unit figures 8 and 9. Now remembering that the 5 
in 58 represents 50 add this to the sum of the unit figures, obtain- 
ing 67 for the sum of 58 and 9 

(a) ib) (c) (d) 

58 9 58 9 

_9 58 _9 58 

17 17 67 67 

50 50 

67 67 

Now notice that the same result may be obtained with less 
figures, in fact mentally, from the following considerations: The 
sum of any two digits cannot exceed 18, since the largest digit is 
9, and 9 + 9 = 18; 18 = 10 + 8 or 1 ten and 8 units. Hence, 
when two numbers are added as above, and the sum of the unit 
figures is 10 or a greater number, write the unit figure of the sum 
of the unit figures and carry the 1 ten to the next left-hand column 
and add it mentally to the figure or figures in that column. Thus, 
in (c) above, 9 and 8 is 17; write 7 and carry 1, which added to 
5 makes 6, and 58 + 9 = 67. In (d), 8 and 9 is 17; write 7 and 
carry 1, which added to 5 makes 6. 

35. The student will find it greatly to his advantage to practice 
adding single digits to numbers less than 100 at every convenient 
opportunity; let him say, for example, 47 and 8 is 55, 23 and 9 is 
32, 36 and 4 is 40, etc., the entire operation being performed 



14 ARITHMETIC §1 

mentally. As soon as proficiency has been obtained in adding 
simple numbers like these, there will be little trouble in adding 
any number of figures correctly. Note that if the sum of the 
figures in the right-hand, or lower, order does not exceed 9, the 
digit in the next order does not change; but if the sum is 10 or 
more, the digit is increased by 1. 

36. If the larger number contains more than two figures, the 
operation is exactly the same, except that if the digit in the next 
to the lowest order is 9, the digit in the next higher order must 
also be increased by 1. Thus 246 and 5 is 251, since 5 + 6=11, 
and 4 + 1 = 5; 594 and 9 is 603, since 9 and 4 is 13, 1 and 9 is 10, 
and 1 and 5 is 6; 1997 and 8 is 2005, since 8 and 7 is 15, 1 and 9 
is 10, 1 and 9 is 10, 1 and 1 is 2; etc. 

37. To Add any Two Numbers. — The sum of any two numbers 
is found as follows: Suppose the numbers are 57,962 and 9,458. 
Write them so that the figures of the same order stand in the same 
column and draw a line underneath. 

57962 

9458 

67420 

Add the figures in the right-hand column first, obtaining 10; write 
the and carry the 1. Add the 1 to the 5 in the next column 
(making 6), and then add this result to the 6 aboVe it, obtaining 
12; write the 2 and carry the 1. Add 1 to the 4 in the next column 
(making 5), and add this sum to the 9 above it, obtaining 14; 
write the 4 and carry the 1. Add 1 to the 9 in the next column 
(making 10), and add this sum to 7 above it, obtaining 17; write 
the 7 and carry the 1. The 1 added to the 5 in the next column 
gives 6, which is written as shown. 

38. The procedure is exactly the same when either or both the 
numbers contain decimals. In this case the easiest way to get the 
figures of the same order under each other is to place the decimal 
points under each other in the same column. 

26.943 0.434 87.51 

5.71 752.933 29.492 

32.653 753.367 117.002 

In all three of the cases, it will be noted that units are placed 
under units, tens under tens, tenths under tenths, etc., and that 



§1 ADDITION AND SUBTRACTION 15 

the decimal point in the result is also directly under the decimal 
points in the numbers added. 

39. To Add more than Two Numbers. — 

Example. — Find the sum of 8, 4, 6, 9, 7 and 3. 

SOLITTION. — 8 

4 
6 
9 

7 
_3 

37 Ans. 
Explanation. — Beginning at the bottom of the column, add the first 
two numbers; to the sum, add the third number; to this sum, add the fourth 
number; etc. Thus, 3 and 7 is 10; 10 and 9 is 19; 19 and 6 is 25; 25 and 4 is 
29; 29 and 8 is 37. After practicing addition in this manner, until a certain 
degree of proficiency has been attained, the reader should abbreviate the 
process by naming only the sums; thus, 3, 10, 19, 25, 29, 37. This will 
greatly increase his speed in adding. 



EXAMPLES 

Find the sum of the following sets of numbers. 

(1) 3+5 + 1+9 + 7 = ? Ans. 25. 

(2) 4 + 2 + 6+8 = ? Ans. 20. 

(3) 3+4+6+8 + 7 = ? Ans. 28. 

(4) 5+8 + 1+6 + 9+4 = ? Ans. 33. 



40. When the numbers contain more than one, figure use the 
following rule: 

Rule. — I. Write the numbers so that the figures of the same order 
will form columns having units under units, tens under tens, tenth 
under tenths, etc., and draw a line under the bottom row of figures. 

II. Beginning with the right-hand column, add the digits in that 
column, and write the unit figure of the sum under the line and in the 
column that was added. Add the tens figure of the sum to the digits 
in the next column to the left, add this column and write the unit 
figure of the sum under the line in the column added. Add the tens 
figure of the sum (if any) to the digits in the third column, proceeding 
in this manner until the addition is finished, which will be when all 
the columns have been added. 

III. If a cipher (0) occurs anywhere, disregard it, since it does 
not affect the sum. 



16 ARITHMETIC §1 

Adding the tens figure of the sum of the digits in any column to 
the digits in the next column to the left is called carrying. 
Example.— 236 + 109 + 871 + 52 + 467 + 696 = ? 
Solution. — 236 

109 
871 
52 
467 
696 
2431 Ans. 
Explanation. — Arrange the numbers as shown, with units under units, 
tens under tens, etc. Beginning with the right-hand column, say 6, 13, 15, 
16, 25, 31; write the 1 under the line and in the column just added, and 
carry 3 to the next column. Say 3, 12, 18, 23, 30, 33; write 3 and carry 3 
to add to the digits in the next column. Say 3, 9, 13, 21, 22, 24. As there 
are no more columns to add, write the last sum, 24, as shown. The sum of 
all the numbers is 2431. 

It is always a good plan to write the abbreviation Ans. (which 
means answer) after the final result has been obtained. 

41. If some of the numbers contain decimals, they are added 
according to the same rule. Arranging the numbers with units 
under units, etc., brings the decimal points under one another 
so that they all stand in the same column ; hence, I of the rule in 
Art. 40 might be changed to read: arrange the numbers so that 
the decimal points stand in the same column. The rest of the 
rule requires no change. 

Example.— Add the following numbers : 403.7819, 21.875, 5.2742, 369.92, 
and 7923.917. 

Solution.— 403.7819 

21.875 
5.2742 
369 . 92 
7923.917 
8724.7681 Ans. 
Explanation. — Arranging the numbers so that the decimal points stand 
in the same column, say 2, 11, and write 1 and carry 1. Say 1, 8, 12, 17, 18; 
write 8 and carry 1. Say 1, 2, 4, 11, 18, 26; write 6 and carry 2. Say 2, 11, 
20, 22, 30, 37; write 7 and carry 3, also writing the decimal point before the 
7. Say 3, 6, 15, 20, 21, 24; write 4 and carry 2. Say 2, 4, 10, 12; write 2 
and carry 1. Say 1, 10, 13, 17; write 7 and carry 1. Say 1, 8, and write 8. 
The entire sum is 8724.7681. 

42. In bookkeeping, and when making out bills, business state- 
ments, etc., the decimal point is usually omitted, a vertical line 
being used in its place. This practice tends to prevent mistakes 



§1 ADDITION AND SUBTRACTION 17 

and saves the writing of decimal points, the vertical line sepa- 
rating the dollars from the cents. 

Example.— Add $66.72, $243.27, $127.85, $37.40, and $101.32. 
Solution.— $ 66 72 

243 27 
127 85 
37 40 
10132 
$576 56 Ans. 

Explanation. — Here the vertical line is used instead of the decimal 
points. The addition is performed in the usual manner, and the sum is 
found to be $576.56. 

43. Checking Results.- — -In business and in engineering, it is of 
the utmost importance that the final result be correct. While 
mistakes or blunders are liable to occur, even when the greatest care 
is taken, they must be overcome by some method of detecting 
them, and corrections must be made before the final result is 
accepted. In checking the work of addition, the various num- 
bers may be added again, but if this is done immediately after the 
first adding, the same mistake is likely to be made again. A 
better way is to add the several columns downward; this causes 
a change in the sequence of the digits added, and tends to prevent 
a second mistake like the first. If the same result is obtained 
when adding down as was obtained when adding up, the work is 
presumed to be correct; but if a different result is obtained, repeat 
the work until the same result is obtained by adding both ways. 
This practice of testing the work to see if it is correct is called 
checking, and the method used in checking is called a check. The 
reader should apply the check by adding down to the two examples 
of the preceding article. 



EXAMPLES 



(1) 26.48 + 360.72 + 54.068 + 7.205 + 509.045 = ? Ans. 957.518. 

(2) 19681 + 89320 + 20358 + 33146 + 7508 = ? Ans. 170013. 

(3) 5818 + 8726 + 6791 + 9809 + 12463 + 752 = ? Ans. 44359. 

(4) 6.317 + 49.42 + 17.718 + 5.801 + 83.77 = ? Ans. 163.026. 

(5) The weights of seven bags of alum are 292 pounds, 305 pounds, 301 
pounds, 298 pounds, 297 pounds, 307 pounds, and 303 pounds; what is their 
total weight? Ans. 2103 pounds. 

(6) What is the weight of six rolls of paper, if the rolls weigh 957 pounds, 
1048 pounds, 1075 pounds, 993 pounds, 986 pounds, and 979 pounds? 

Ans. 6038 pounds. 



18 ARITHMETIC §1 

(7) Paid the following amounts for bales of rags: $18.82, $18.94, $19.11, 
$19.20, $18.85, $18.97. What was the total amount paid? Ans. $113.89. 

(8) Six barrels of rosin were placed on an elevator; if the barrels weighed 
389 pounds, 411 pounds, 395 pounds, 399 pounds, 408 pounds, and 406 
pounds, how much must the elevator raise, if it carries in addition to the 
rosin a man weighing 168 pounds? Ans. 2576 pounds. 



SUBTRACTION 

44. Subtraction means taking away. In arithmetic, subtrac- 
tion is the process of taking one number from another number, or 
it is the process of finding how much greater one number is than 
another. If a person has 10 cents and spends 4 cents, he will 
have 6 cents left. Here 4 cents are taken from 10 cents, and 6 
cents are left. The arithmetical operation of taking 4 cents 
from 10 cents is called subtraction. Again, how much greater is 
10 cents than 4 cents? If 6 cents are added to 4 cents, the sum 
is 10 cents; hence, 10 cents is 6 cents greater than 4 cents. The 
same result will be obtained by subtracting 4 cents from 10 cents. 

45. The sign of subtraction is — , a short horizontal line; it is 
read minus, and means less; it indicates that the number following 
it is to be subtracted from the number preceding it. Thus, 10 — 
4 = 6; here 4 is to be subtracted from 10, the result being 6; the 
expression may be read either as 10 minus 4 equals 6 or as 10 less 
4 equals 6. The longer expression, 4 subtracted from 10 equals 
6, is also correct. 

46. The number subtracted is called the subtrahend, and the 
number from which the subtrahend is subtracted is called the 
minuend; the result is called the remainder or difference. When 
it is not desired to be specific regarding the order of the numbers, 
the word difference is generally used ; thus, the difference between 
4 and 10 or the difference between 10 and 4 is 6. But when the 
order of the numbers is definitely stated, as 10 minus 4 is 6, 6 
is called the remainder, it being what is left after taking 4 from 10. 
In an expression like 15 — 9 = 6, 15 is the minuend, 9 is the 
subtrahend, and 6 is the remainder. 

Before explaining the general process of subtraction, the two 
following principles or laws should be thoroughly understood: 

47. Principle I. — If the same number he added to both minuend 
and subtrahend, the remainder is unchanged. For example, 15 — 
9 = 6; if now, 10 be added to both 15 and 9, they become 25 



§1 ADDITION AND SUBTRACTION 19 

and 19 respectively, and 25 — 19 = 6, the same remainder as 
before. The reason is evident; the 10 added to the subtrahend is 
subtracted from the 10 added to the minuend, thus leaving the 
minuend and subtrahend the same as before the 10 was added. 
A like result will be obtained, no matter what number is added; 
thus, adding 7 to 15 and 9, these numbers become 22 and 16, and 
22 — 16 = 6, as before. 

48. Principle II. — If the remainder he added to the subtrahend, 
the sum will he the minuend. For instance, if a stick is 13 feet 
long and 5 feet are cut off-, the length of the stick is then 8 feet. 
Here 13 — 5 = 8, and 5 is the subtrahend and 8 is the remainder. 
But when the two parts of the stick are placed together, they 
must be equal in length to the original stick; that is, 5 + 8 = 13, 
or subtrahend + remainder = minuend. Hence, instead of 
saying 5 from 13 leaves 8, it will be equally proper to say "what 
number added to 5 will make 13?" this number is 8; therefore, 
13 - 5 = 8, because 5 + 8 = 13. 

49. The reader will find it greatly to his advantage to reverse 
the addition table of Art. 33. Thus, 9 and 6 is 15; 15 less 9 is 6, 
and 15 less 6 is 9. Again, 8 and 5 is 13; 13 less 8 is 5, and 13 less 
5 is 8. He should practice in this manner both addition and 
subtraction in spare moments, until the result presents itself 
instantly as soon as the numbers are named. 

50. If the right-hand figure of the minuend is smaller than the 
digit to be subtracted, add 10 to it, subtract, and then subtract 1 
from the next left-hand figure of the minuend; thus, 34 — 8 = 
26. Here 10 is added to 4, making 14; 8 from 14 leaves 6; then 
subtracting 1 from 3, the remainder is 2. This is in accordance 
with Principle I, Art. 47, since 10 is added to the minuend (really 
making it 44, but expressed as 30 + 14) and 10 is added to the 
subtrahend (really making it 18). The result is correct; since, 
by Principle II, 8 + 26 = 34; also 18 -|- 26 = 44. It is thus 
seen that 36 - 9 = 27, 43 - 7 = 36, 81 - 4 = 77, etc. The 
reader should practice these combinations also. 

61. Bearing the foregoing in mind, the following is the rule for 
subtraction : 

Rule I. — Place the numbers as in addition, with the subtrahend 
under the minuend, and with units under units, tens under tens, 
etc. or with the decimal ^points in the same column. 

II. Beginning with the right-hand digit of the subtrahend, sub- 



20 ARITHMETIC §1 

tract it from the digit above it in the minuend; do the same with the 
next digit or figure to the left, proceeding in this manner until all 
the figures in the subtrahend have been subtracted from the figures 
above them in the minuend. 

III. If a figure in the minuend is a cipher or is smaller in value 
than the figure under it in the subtrahend, add 10 to it before sub- 
tracting, and then carry 1 to the next figure of the subtrahend before 
subtracting it from the figure above. 

IV. Having found the remainder, add it to the subtrahend, figure 
by figure, and if the sum equals the minuend, the remainder is very 
probably correct. This last operation is a check. 

Example 1.— From 93875 take 72032. 

Solution.— 93875 

72032 

21843 Ans. 

Explanation. — Writing the numbers as directed in I of the rule, begin 
with the right-hand digit and say 2 from 5 is 3, 3 from 7 is 4, write 8 (since 
from any number leaves the number), 2 from 3 is 1, and 7 from 9 is 2. The 
remainder, therefore, is 21,833. To check this result, add the remainder 
to the subtrahend; thus, 3 and 2 is 5, 4 and 3 is 7, 8, 1 and 2 is 3, and 2 and 
7 is 9. The sum being the same as the minuend, the work is very probably 
correct. 

Example 2.— From $1437.50 take $918.27. 

Solution.— $1437.50 

918.27 
$ 519.23 Ans. 

Explanation. — Since 7 cannot be subtracted from 0, add 10 to 0, obtain- 
ing 10; then 7 from 10 is 3. Carry 1 to 2 making it 3, and 3 from 5 is 2. 
Write the decimal point. Then, since 8 cannot be taken from 7, add 10, 
making 17; and 8 from 17 is 9. Carry 1 to 1 making it 2, and 2 from 3 is 1. 
Lastly, 9 from 14 is 5. If the remainder be added to the subtrahend, the 
sum is the minuend. Hence, the correct remainder is $519.23. 

Example 3. — There are four piles of papers; the first pile contains 986 
sheets, the second pile contains 753 sheets, the third pile contains 875 sheets, 
and the fourth pile contains 1038 sheets; they were all placed in one pile, 
and 2369 sheets were taken away; how many sheets were left? 

Solution. — The first step is to find how many sheets 
there were all together. This found by adding the "°" sheets 

number of sheets in each pile, the sum or total being '^'^ 

3652 sheets. The next step is to subtract from this °'^ 

sum the number of sheets that were taken away, the l^oo 
remainder thus found being 1283 sheets. To check, 3652 sheets 
add down, then add the remainder to the subtrahend, 2369 
obtaining 3652, the minuend. Since the work checks 1283 sheets .Ans. 
in both cases, it is probably correct. 



§1 ADDITION AND SUBTRACTION 21 

Example 4.— From 19.125 take 9.3125. 

Solution.— 19.1250 

9.3125 
9.8125 Ans. 

Explanation. — There is no figure in the minuend above the right-hand 
figure of the subtrahend; but since the subtrahend contains a decimal, 
it is allowable (for reasons that will be explained later) to annex ciphers 
to the minuend until it contains the same number of figures in the decimal 
part that there are in the subtrahend. The subtraction is then performed 
in the regular manner. Should there be more figures in the decimal part of 
the minuend than in the decimal part of the subtrahend, annex ciphers to 
the subtrahend until the decimal parts of the subtrahend and minuend both 
contain the same number of figures, before subtracting. 

52. Although subtraction is a much simpler and easier opera- 
tion than addition, it is probable that more mistakes are made in 
subtraction than in addition, the reason being that the operator 
fails to check by "adding back" and is more inclined to be care- 
less. It is well not to attempt to work too fast at first; try to 
secure accuracy, and speed will come with practice. This advice 
applies to every process of computation. 



EXAMPLES 

(1) From 417.23 take 273.58. Ans. 143.65. 

(2) From 54.375 take 48.65625. Ans. 5.71875. 

(3) From 484814 take 258177. . Ans. 226637. 

(4) From 1002070 take 304098. Ans. 697972. 

(5) 2033.4238 - 1792.3917 = ? Ans. 241.0321. 

(6) From a pile of pulp weighing 14,253 tons 9,468 tons were sold; how 
much pulp remained in the pile? Ans. 4,785 tons. 

(7) Three cars were receiv^ed containing 58,024 pounds of sulphur; when 
the stock was cleaned up, mill figures showed that 55,638 pounds had been 
used; how much had been lost in handling? Ans. 2,386 pounds. 

(8) To a pile of pulp weighing 4,826 tons, 527 tons were added; how much 
was left after 3,962 tons had been sold? Ans. 1,391 tons. 

(9) A bleach liquor tank contained 925 gallons; how much remained after 
146 gallons had been used? Ans. 779 gallons. 



22 ARITHMETIC §1 

MULTIPLICATION AND DIVISION 



MULTIPLICATION 

63. In arithmetic, multiplication may be defined as a short 

method of adding (or finding the sum of) several equal numbers. 

For instance, suppose there are 6 books, 

(") each book containing 1892 pages; how 

1892^^^^^ ^^^^ P^^®^ ^^^ ^^ *^^ ^ books? To 

jgg2 find the total number, 1892 may be 

1892 written six times as shown at (a), and the 

1892 sum found, which is 11352 pages. By 

^^^^ using the process of multiplication, how- 

11352 pages Ans. ever, the work will be arranged as shown 

,,. at (6), and 1892 is then multiplied by 6, 

jgg2 the result being the same as before. 

6 Note the great saving in figures even in 

11352 pages Ans. this simple case. Suppose there had 

been, say, 576 books each containing 

1892 pages; it would then be practically impossible to find the 

total number of pages by the method shown at (a), but the total 

can easily be found by multiplication. 

54. The number multiplied (1892 in Art. 53) is called the multi- 
plicand; the number used to multiply the multiplicand (and 
which shows how many times the multiplicand is to be added) is 
called the multiplier (this is 6 in Art. 53) ; the result of the opera- 
tion of multiplication is called the product (in Art. 53, 11352 is 
the product). 

When it is not desired to be specific and make a special dis- 
tinction between the multiplicand and multiplier, these two num- 
bers are called factors; in Art. 53, 1892 and 6 are factors of 11352, 
their product. 

55. The sign of multiplication is an oblique (St. Andrew's) 
cross, which is written between the factors; it is read times or 
multiplied by. Thus, 6 X 8 = 48 is read either as six times 
eight equals forty-eight or as six multiplied by eight equals forty- 
eight; here 6 and 8 are factors, 6 being the multiplier in the first 
case and 8 being the multiplier in the second case. 

56. Insofar as the product is concerned, it makes no difference 
which of two factors is considered as the multiplier, since 6 times 



MULTIPLICATION AND DIVISION 



23 



8 is the same, for instance, as 8 times 6, the product in both cases 
being 48; this may be expressed mathematically as 6 X 8 = 8 
X 6 = 48. 

57. Before one can multiply, it is necessary that he memorize 
the multiplication table. This may take a little time, but it is 
absolutely necessary if the reader is to be successful in this 
subject. 

MULTIPLICATION TABLE 



1 times 


1 


s 


1 


2 times 


1 


s 


2 


3 times 1 is 


3 


4 times 1 is 4 


1 times 


2 


s 


2 


2 times 


2 


s 


4 


3 times 2 is 


6 


4 times 2 is 8 


1 times 


3 


s 


3 


2 times 


3 


s 


6 


3 times 3 is 


9 


4 times 3 is 12 


1 times 


4 


s 


4 


2 times 


4 


s 


8 


3 times 4 is 


12 


4 times 4 is 16 


1 times 


5 


s 


5 


2 times 


5 


s 


10 


3 times 5 is 


15 


4 times 5 is 20 


1 times 


6 


s 


6 


2 times 


6 


s 


12 


3 times 6 is 


18 


4 times 6 is 24 


1 times 


7 


s 


7 


2 times 


7 


s 


14 


3 times 7 is 


21 


4 times 7 is 28 


1 times 


8 


s 


8 


2 times 


8 


s 


16 


3 times 8 is 


24 


4 times 7 is 32 


1 times 


9 


s 


9 


2 times 


9 


s 


18 


3 times 9 is 


27 


4 times 9 is 36 


1 times 


10 


s 


10 


2 times 


10 


s 


20 


3 times 10 is 


30 


4 times 10 is 40 


1 times 


11 


s 


11 


2 times 


11 


s 


22 


3 times 11 is 


33 


4 times 11 is 44 


1 times 


12 


s 


12 


2 times 


12 


s 


24 


3 times 12 is 


36 


4 times 12 is 48 


5 times 


1 


s 


5 


6 times 


1 


s 


6 


7 times 1 is 


7 


8 times 1 is 8 


5 times 


2 


s 


10 


6 times 


2 


s 


12 


7 times 2 is 


14 


8 times 2 is 16 


5 times 


3 


s 


15 


6 times 


3 


s 


18 


7 times 3 is 


21 


8 times 3 is 24 


5 times 


4 


s 


20 


6 times 


4 


s 


24 


7 times 4 is 


28 


8 times 4 is 32 


5 times 


5 


s 


25 


6 times 


5 


s 


30 


7 times 5 is 


35 


8 times 5 is 40 


5 times 


6 


s 


30 


6 times 


6 


s 


36 


7 times 6 is 


42 


8 times 6 is 48 


5 times 


7 


s 


35 


6 times 


7 


s 


42 


7 times 7 is 


49 


8 times 7 is 56 


5 times 


8 


s 


40 


6 times 


8 


s 


48 


7 times 8 is 


56 


8 times 8 is 64 


5 times 


9 


s 


45 


6 times 


9 


s 


54 


7 times 9 is 


63 


8 times 9 is 72 


5 times 


10 


s 


50 


6 times 


10 


s 


60 


7 times 10 is 


70 


8 times 10 is 80 


5 times 


11 


s 


55 


6 times 


11 


s 


66 


7 times 11 is 


77 


8 times 11 is 88 


5 times 


12 


s 


60 


6 times 


12 


s 


72 


7 times 12 is 


84 


8 times 12 is 96 


9 times 


1 


s 


9 


10 times 


1 


3 


10 


11 times 1 is 


11 


12 times 1 is 12 


9 times 


2 


s 


18 


10 times 


2 


s 


20 


11 times 2 is 


22 


12 times 2 is 24 


9 times 


3 


s 


27 


10 times 


3 


s 


30 


11 times 3 is 


33 


12 times 3 is 36 


9 times 


4 


s 


36 


10 times 


4 


s 


40 


11 times 4 is 


44 


12 times 4 is 48 


9 times 


5 


s 


45 


10 times 


5 


s 


50 


11 times 5 is 


55 


12 times 5 is 60 


9 times 


6 


s 


54 


10 times 


6 


s 


60 


1 1 times 6 is 


66 


12 times 6 is 72 


9 times 


7 


s 


63 


10 times 


7 


s 


70 


11 times 7 is 


77 


12 times 7 is 84 


9 times 


8 


s 


72 


10 times 


8 


s 


80 


11 times 8 is 


88 


12 times 8 is 96 


9 times 


9 


s 


81 


10 times 


9 


s 


90 


11 times 9 is 


99 


12 times 9 is 108 


9 times 


10 


IS 


90 


10 times 


10 


s 


100 


11 times 10 is 


110 


12 times 10 is 120 


9 times 


11 


IS 


99 


10 times 


11 


s 


110 


11 times 11 is 


121 


12 times 11 is 132 


9 times 


12 


s 


108 


10 times 


12 


IS 


120 


11 times 12 is 


132 


12 times 12 is 144 



The product of any number and is (9 X = 0, 167 X = 
0, etc.) ; this is evident, since the sum of any number of O's cannot 
make an integer. 

It will be noted by reference to the table that the product of any 
number and 1 is the number itself; thus, 409 X 1 = 409, since 



24 ARITHMETIC §1 

409 X 1 is 409 I's, which is, of course, 409. Note, also, that the 
product of any number and 10 is the number itself with a cipher 
(0) added at the right; thus, 7 X 10 = 70, 526 X 10 = 5260, 
etc. Note, again, that the product of 11 and any number not 
greater than 9 is the number repeated; thus, 3X11 = 33, 6X 
11 = 66, 9 X 11 = 99, etc. 

The reader should repeat the different parts of the table to 
himself at odd times until it becomes so firmly impressed on his 
memory that as soon as any two numbers are named, their pro- 
duct will instinctively name itself. 

58. To Multiply any Number by a Single Digit. — Multiply 

31415927 by 7. Here 31415927 is the 

_ multiplicand and 7, the multiplier, is 

omnn^cf^ A written under the right-hand figure of the 

219911489 Ans. . . ° . * 

multiplicand. Draw a line under the 
factors, as shown, and multiply the right-hand figure of the 
multiplicand by 7 the multiplier, obtaining 7X7 = 49. Write 
the unit figure of the product under the unit figure of the multipli- 
cand, and carry 4, the tens figure of this product. Then say 7 
times 2 is 14, to which add 4, the figure carried, making 18; write 
8 as shown and carry 1. Say 7 times 9 is 63, add the 1 carried, 
making 64; write 4 and carry 6. Say 7 times 5 is 35 and 6 (the 
figure carried) is 41; write 1 and carry 4. Say 7 times 1 is 7 and 
4 (the figure carried) is 11; write 1 and carry 1. Say 7 times 4 is 
28 and 1 is 29; write 9 and carry 2. Say 7 times 1 is 7 and 2 is 9; 
write 9 and there is nothing to carry. Lastly, say 7 times 3 is 
21, v/hich write. Every figure in the multiplicand has now been 
multiplied by the multiplier, 7, and the product is 219,911,489. 
Had the multiplier been 70, 700, 7000, etc., the process would 
have been exactly the same, except that after the product was 
found, as many ciphers would have been annexed to the product 
as there were ciphers to the right of the right-hand digit of the 
multiplier; thus, 31,415,927 X 7000 = 
31415927^^^ 219,911,489,000. The work is shown in 

219911489000 Ans. *^^ margin. First multiply by 7 as 

before; then to the product, annex three 
ciphers, because there are three ciphers to the right of the right- 
hand digit of the multiplier . 

59. To Multiply When the Multiplier Contains Two or More 
Digits. — Place the multiplier under the multiplicand, with the 



§1 MULTIPLICATION AND DIVISION 25 

right-hand digit of the multiplier under the right-hand digit of 
the multiplicand. Multiply by the right-hand digit of the multi- 
plier and write the product figure by figure under the multiplier, 

as shown in the margin. This result is 
428095 called the first partial product. Note that 

after multiplying 9 by 2 there is 1 to carry; 



1712380 *^®^ ^^y ^ times naught is naught and 1 is 

29966650 Ij which write as shown. Now multiply by 

1284285 the next figure to the left of 2, in this case 

15857494990 Ans. 4. Say 4 times 5 is 20; write the cipher 
one place to the left of the right-hand figure 
of the first partial product, thus bringing the cipher under the 
figure multiphed by. Continue the multiphcation by 4, obtain- 
ing 1712380, which is called the second partial product. The third 
figure to be used as a multiplier is 0, and since any number 
multiphed by is 0, write a cipher one place to the left of the 
right-hand figure of the second partial product, which brings it 
directly under the cipher in the multiplier. Now multiply by 
the next figure, 7, of the multiplier. Say 7 thnes 5 is 35, write 
5 alongside the cipher and carry 3; this brings the 5 under the 
figure used as a multiplier, and makes the third row of figures 
29966650, the third partial product, which is equal to 428095 X 
70. Finally, multiply by 3, the left-hand digit of the multiplier, 
and the result is the fourth partial product, the right-hand figure 
of which is written under 3, the number multiplied by. Now 
adding the four partial products, the sum is 15,857,494,990, 
which is the entire product, or the result sought. 

If there are ciphers to the right of the multiplicand or multiplier 
or both, pay no attention to them until after the product has 
been found as just described. Then annex to the entire product 
as many ciphers as there are ciphers to the right of either or both 

factors. For instance, to multiply 

205000 526700 by 205000, arrange as shown in 

26335 — *^^ margin, with the right-hand digits 

105340 o^ *^6 multipHcand and multiplier under 

107973500000 Ans. ^ach other. Multiply 5267 by 205, the 

product being 1079735; there are two 

ciphers to the right of one factor and three to the right of the 

other factor; hence, annex 2 + 3 = 5 ciphers to the right of the 

entire product, which is thus found to be 107,973,500,000. 



26 ARITHMETIC §1 

60. Rule. — Write the multiplier under the multiplicand with the 
right-hand digits under each other. Beginning with the right-hand 
digit of the multiplier, and proceeding to the left, multiply the upper 
factor by each figure of the loiver factor, or multiplier, writing the 
right-hand figure of each partial product under the figure used as a 
multiplier. Then add the partial products, and the sum will be the 
entire product. 

61. Check for Multiplication. — The best way to check multi- 
pHcation is to employ the process called "casting out nines." 
This consists in dividing (the operation of dividing will be con- 
sidered in the next article) the two factors by 9, multiplying the 
remainders, and if the product is greater than 9, divide that by 9; 
note the remainder. Then divide the entire product by 9, and 
if the remainder is the same as that first obtained, the work is 
very probably correct. If the two remainders differ, however, 
then the work is wrong, some mistake having been made. In- 
stead of dividing by 9, the remainder may be found by adding the 
digits; if the sum is greater than 10, add the digits of the sum, pro- 
ceeding in this manner until a single digit has been found, which 
will be the remainder when the number is divided by 9. Thus, 
consider the number 7854; the sum of the digits is 7 + 8 + 5 + 4 
= 24; 2 + 4 = 6, and 6 is the remainder when 7854 is divided by 9. 

Applying this check to the first example of Art. 59, the sum of 
the digits in the multipHcand is 4 + 2 + 8 + 9 + 5- 28, 2+8 
= 10, and 1 + = 1; the sum of the digits in the multiplier is 
3 + 7 + 4 + 2 = 16, and 1 + 6 = 7; then 1X7 = 7. The 
sum of the digits in the entire product is 1 + 5 + 8 + 5 + 7 + 4 + 9 
+ 4 + 9 + 9 = 61, and 6 + 1=7. Since the remainders are 
both 7, the work is very probably correct. When adding the 
digits in this manner, it is not necessary to add any 9's; thus, in 
the foregoing, the remainder for the multiplicand is 4 + 2 + 8 + 

5 = 19 or 1, and the remainder for the product is 1 + 5 + 8 + 5 
+ 7 + 4 + 4 = 34, and 3 + 4 = 7, both results being the same 
as obtained before. 

Applying this check to the second example of Art. 59, 5 + 2 + 

6 + 7 = 20, or 2; 2 + 5 = 7; 2 X 7 = 14, and 1 + 4 = 5; 
1 + 7 + 7 + 3 + 5 = 23, and 2 + 3 = 5. Since the remain- 
ders are the same, the work is very probably correct. 

The reader is strongly advised to apply this check in every 
case. 



§1 MULTIPLICATION AND DIVISION 27 

EXAMPLES 

(1) 7854 X 2038 = ? Ans. 16,006,452. 

(2) 230258 X 90057 = ? Ans. 20,736,344,706. 

(3) 31831 X 31416 = ? A7is. 1,000,002,696. 

(4) 543836 X 4688 = ? Ans. 2,549,503,168. 

(5) 197527 X 98743 = ? Ans. 19,504,408,561. 

(6) 295369 X 700405 = ? Ans. 206,877,924,445. 

(7) The average consumption of coal by a mill per year is 28,750 tons; if 
the average cost per ton is $7.00, what is the annual cost to the mill for coal? 

Am. $201,250. 

(8) During a certain period, a mill consumed 686 tons of alum ; if the price 
paid for the alum was $49.00 per ton, how much was the total amount paid 
for the alum? Ans. $33,614. 

(9) What is the value of 13,908 cords of wood at $11.00 per cord? 

Ans. $152,988. 

(10) How much must be paid for 12 cans of dye stuff, if each can weighs 
47 pounds and the dye stuff is worth $19.00 per pound? Ans. $10,716. 

(11) At different times a certain mill sold paper as follows: 27,848 
pounds at 14 cents per pound; 17,005 pounds at 18 cents per pound; 9,990 
pounds at 19 cents per pound; and 36,476 pounds at 15 cents per pound. 
How much was received from these sales? Ans. $14,329.12. 



DIVISION 

62. Division means a partition, a separating into parts. In 

arithmetic, division is the process of finding how many times 

larger one number is than another; thus, since 24 

11352 ^^ ^ times 4, 24 is 4 times as large as 6 or 6 times as 

1892 large as 4. Division may also be defined as the 

9460 process of separating a number into a required 

1892 number of equal parts; thus, 24 may be separated 

7568 into 6 equal parts of 4 each or 4 equal parts of 6 each. 

1^92 Just as multiplication is a short process or 

5676 method of adding equal numbers so division is a 

short process or method of subtracting continuously 

^^^ until the remainder is or less than the subtra- 

— — hend. For example, referring to Art. 53, suppose 

1892 there are 11,352 pages in a certain number of books 

^^ each containing 1892 pages, and it is required to 

find the number of books. The work might be 

^^<^ done as shown at (a) in the margin, subtracting 

11352 ^^^2 from 11352, then subtracting 1892 from the 

Q remainder, continuing this process until the 

remainder becomes or less than the subtrahend. 



28 ARITHMETIC §1 

In this case, the remainder is 0; and as the subtraction was 
performed 6 times, there are 6 books. By the process of divi- 
sion, as shown at (b), 1892 is contained 6 times in 11352, because 
1892 X 6 = 11352. Note the great saving of j&gures in the 
second method. Had it been known that the total number of 
pages was 11,352, the number of books was 6, and it were 
required to find the number of pages in each book, it would be 
necessary to subtract 6, by the method shown at (a), 1892 times, 
which is practically impossible. 

In multiplication, the object to be attained is to find the 
product of two numbers (factors); in division, the object is to 
divide a number into two factors, one of them being given. 

63. The number that is to be divided into two factors is called 
the dividend; the given factor, which is divided into the dividend, 
is called the divisor; the other factor, which is obtained by divid- 
ing the dividend by the divisor, is called the quotient; anything 
that is left over after the division has been performed is called the 
remainder. In (b), Art. 62, 11,352 is the dividend, 1892 is the 
divisor, 6 is the quotient, and the remainder is 0. Whenever the 
remainder is 0, the division is said to be exact. 

64. There are several signs for division, the principal one being 
a colon ( :) or a colon with a short horizontal line between the dots 
(-^). When either of these two signs occurs between two 
numbers, it means that the number on the left is to be divided by 
the number on the right; thus, 24 -^ 6 means that 24 is to be 
divided by 6, 24 being the dividend and 6 the divisor. In some 
cases, a vertical line is used in place of the regular sign of division; 
thus, 24 1 6. The vertical line is seldom used between two 
numbers; it is most frequently used when the product of several 

84 numbers is to be divided by the product of 

112 several other numbers; thus, the product of 124, 

49, and 75 divided by the product of 84 and 112 
may be indicated as shown in the margin, the product of the 
numbers to the left of the vertical line being the dividend, and the 
product of the numbers to the right being the divisor. Most 
commonly, however, in cases of this kind, a horizontal line 
is used, the number or numbers above the line being divided 

24 . 

by the number or numbers below it; thus, -^ is read 24 over 6, 

and means 24 divided by 6; also, — qa v 1 1 9 — means that the 



124 
49 

75 



§1 MULTIPLICATION AND DIVISION 29 

product of 124, 49, and 75 is to be divided by the product of 84 
and 112. An inclined line is also frequently used; thus 24/6 
means 24 divided by 6. 

65. Short Division. — When the divisor is not greater than 12, it 

is customary to employ what is called short 

division. The process is best illustrated 

8)5 463^863^2 by an example. For instance, 543,832:8 = ? 

6 7 9 7 9 Write the divisor to the left of the dividend 

with a curved line between, as shown in the 

margin, and draw a straight line under the 

dividend. Since 8 is greater than 5, the left-hand digit of the 

dividend, consider the first two figures, 54. Now find what 

number multiplied by 8 will make 54 or whose product subtracted 

from 54 will be less than 8; since 8 X 6 = 48 and 8 X 7 = 56, 

this number is 6, and 54 — 48 = 6. Write 6 under 54 for the 

first figure of the quotient, and also write 6, the remainder, above 

and to the right of 4, as shown. This last 6 belongs to the same 

order as the 4 in 54 and the 3 that follows 4 is of the next lower 

order; hence, combine the 6 and 3, and call the number 63. Now 

find what number multiplied by 8 will make 63 or whose product 

subtracted from 63 will be less than 8; this number is 7, since 8 X 

7 = 56 and 63 — 56 = 7. Write 7 under 3 for the second figure 
of the quotient, and also write 7, the remainder, as shown. The 
next number to be divided is 78, the 7 being prefixed to the fourth 
figure of the dividend, which is the figure of the next lower order. 
Here 8 X 9 = 72, and 78 - 72 = 6; write 9 under 8 for the third 
figure of the quotient and write 6, the remainder, as shown. 
Prefixing 6 to 3, the next figure of the dividend, 8 X 7 = 56, 63 
— 56 = 7; hence, write 7 for the fourth figure of the quotient and 
write 7, the remainder, as shown. Finally, 8 X 9 = 72, and 
there is no remainder; hence, write 9 for the fifth figure of the 
quotient. As there are no more figures in the dividend, the quo- 
tient sought is 67,979. That this is correct may be proved by 
multiplying the quotient by the divisor, the result being 67979 X 

8 = 543832, the dividend. 

In practice, the remainders would not be written in the manner 
indicated in the foregoing^ — they would be simply carried men- 
tally. The process would then be as follows : 8 into 54, 6 times 
and 6 over; 8 into 63, 7 times and 7 over; 8 into 78, 9 times and 
6 over; 8 into 63, 7 times and 7 over; 8 into 72, 9 times and no re- 



30 ARITHMETIC §1 

mainder. Suppose it were required to divide the above number, 
543,832 by 9. Say 9 into 54, 6 times and 

^ nothing over; 9 into 3 no times; 9 into 38, 

60425+ 7 rem. ^ ^.^^^ ^^^ 2 over; 9 into 23, 2 times and 

5 over; 9 into 52, 5 times and 7 over. Since there are no more 
figures in the dividend, the quotient is 60,425 and 7 remainder. 
That by result is correct may be proved by multiplying the 
quotient by the divisor and adding the remainder to the product; 
thus, 60425 X 9 = 543825, and 543825 + 7 = 543832. Now 
note that the remainder is the same as that obtained in Art. 61 by 
adding the digits; thus 5 + 4 + 3 + 8 + 3 + 2 = 25, and 2 + 5 
= 7. 

For reasons that will be explained later, it is customary to write 
the remainder over the divisor, with a line between, and annex 
this expression to the quotient. In the last example, 543832 -j- 9 
= 60425^. Ans. This last expression may be read sixty 
thousand four-hundred-twenty-five and seven over nine. 



EXAMPLES 

(1) 197527 -4- 11 = ? Ans. 17,957. 

(2) 527324 ^ 7 = ? Ans. 75,332. 

(3) 900725 -H 6 = ? Ans. 150,120f. 

.^, 49503163 „ A A ioro«Q7 

(4) Tg = ? Ans. 4,125,263i^. 

(5) 1580216/4 = ? Ans 395,054. 

(6) 4350688 ^ 3 = ? Ans. l,450,229i. 

(7) 2072623/10 = ? Ans. 207,262i3^. 

Note. —To divide a number by 10, write all the figures except the last for the quotient; 
the last figure will be the remainder (see example 7, above) . 

(8) Since there are 12 inches in one foot, how many feet are equivalent to 
237 inches? Ans. l^-^^ feet. 

(9) How many nickels are equal in value to $3.45, one nickel being equal 
to 5 cents? Ans. 69 nickels. 

(10) Twelve of anything make a dozen, and twelve dozen make a gross. 
How many dozen balls of twine are in a shipment containing 5076 balls? 
also, how many gross were in the shipment ? Ans. 423 dozen ; 35^ gross. 

(11) There are 8 pints in one gallon; how many gallons are equivalent to 
22,222 pints? Ans. 2777| gallons. 

(12) One yard is equal to three feet; how many yards are contained in 
63360 inches? Ans. 1760 yards. 

(13) A quire of paper contains 24 sheets. How many quires are in a pile 
of 1784 sheets? Ans. 74^^ quires. 



§1 MULTIPLICATION AND DIVISION 31 

66. Long Division. — When the divisor is greater than 12, the 
process called long division is used; this is best explained by an 

example. For instance, what is 
64903358) 386 the quotient when 64,903,358 is 

386 168143Hi ^«s. divided by 386 ? Write the divi- 

2630 sor to the right of the dividend, 

2316 with a line (either straight or 

3143 curved) between, and draw a line 

^Q^^ under the divisor, as shown. 

^^^ Since the divisor contains 3 fig- 

— ures, compare it with the number 

Jg^^ made up of the first 3 figures of 

-r^g the dividend, in this case, 649, 

jj5g which call the first trial dividend. 

"^ Note that 649 is larger than 386, 

which is nearly equal to 400; 
calling it 400, it is seen that 400 is contained in 649, 1 time, and 
1 is thus the first figure of the quotient, which is written under 
the divisor. Now multiply the divisor by 1, the figure of the 
quotient just found, and write the product under the first three 
figures of the dividend, draw a line under it, and subtract, obtain- 
ing a remainder of 263. Annex to this remainder the next figure 
of the dividend, in this case 0, and divide 2630, which is the new, 
or second, trial dividend, by 386, the divisor, which call 400 as be- 
fore, obtaining 6 for the second figure of the quotient. Multiply 
the divisor by 6, the figure of the quotient last found, and 
write the product, 2316, under 2630; subtract as before, obtain- 
ing 314 as a remainder, to which annex the next figure of the 
dividend, 3 in this case, making the new, or third, trial dividend 
3143. To divide 3143 by 386, note that 8 X 400 = 3200, a 
number slightly larger than 3143; but as 386 is smaller than 
400, try 8 for the third figure of the quotient. Multiplying 386 
by 8, the product is 3088, which subtracted from 3143, leaves a 
remainder of 55, to which annex the next figure of the divi- 
dend, in this case 3, making the new, or fourth, trial dividend 
553. The next figure of the quotient is evidently 1, and the next, 
or fifth, trial dividend is 1675. Dividing 1675 by 400, the next 
figure of the quotient is 4; multiplying 386 by 4 and subtracting 
the product from 1675, the remainder is 131, to which annex the 
next (in this case, the last) figure of the dividend, obtaining 1318 
for the new, or sixth, trial dividend. Dividing 1318 by 400, the 



32 ARITHMETIC §1 

next figure of the quotient is 3 ; the remainder after multiplying 
the divisor by 3 is 160. As there are no more figures in the divi- 
dend, the division of 160 by 386 is indicated by writing 386 under 
160 with a line between. The quotient, therefore, is 168,143m. 

In the foregoing, the number 400, which was used in place of 
386 to determine the different figures of the quotient, is called the 
trial divisor. If the second figure of the divisor is 5 or a larger 
digit, increase the first figure of the divisor by 1, use the result 
thus obtained as a trial divisor, and proceed as in short division 
to obtain the next figure of the quotient. But, if the second 
figure of the divisor is less than 5, use the first figure of the divisor 
for a trial divisor. In case there is any doubt as to whether the 
figure of the quotient so obtained is correct, multiply the second 
figure of the divisor by the figure thus obtained in the quotient 
and add the amount to be carried to the product of this figure and 
the first figure of the divisor, comparing the result with the first 
two figures of the trial dividend. Thus, in the foregoing ex- 
ample, to determine whether to try 7 or 8 for the third figure of 
the quotient 8 X 8 = 64, 8 X 3 = 24, and 24 + 6 = 30; since 
30 is smaller than 31, the first two figures of the trial dividend, 
try 8 for the third figure of the quotient. It may be remarked 
that the quotient, 168143^|t, is read one hundred sixty-eigh 
thousand one hundred forty-three and one hundred sixty over 
three hundred eighty-six. 

67. Check for Division. — To check division, cast out 9's from 
the dividend, divisor, quotient, and remainder; the product of 
the remainders for the divisor and quotient plus the remainder 
for the remainder should equal the remainder for the dividend, 
but if not, a mistake has been made. Thus, in the example of 
Art. 66, disregarding the 9's, 6 + 4 + 3 -|- 3 + 5 -F 8 = 29, and 
the remainder is 2 for the dividend. The remainder for the divi- 
sor is 3 + 8 + 6 = 17, and 1 -|-7 = 8; the remainder for the 

1 .9001 S7^r u.r 07 quotient is 1 + 6 + 8 + 1+4 + 3 = 

1529918746)43607 _„ j o i o c +u • j 

Toncoi in -7^0 23, and 2 + 3 = 5; the remamder 

lo0821 35084 ^-^J--^-— Ans 

221708 ^ ^ f°^ ^^® remainder is 1 + 6 = 7; then 

218035 8 X 5 = 40, or 4, and 4 + 7 = 11, 

367374 1 + 1 = 2, the same remainder as 

348856 was found for the dividend; hence, 

185186 the work is probably correct. 

174428 As another example, divide 

10758 1,529,918,746 by 43607. Since 



§1 MULTIPLICATION AND DIVISION 33 

the second figure of the divisor is 3, a digit smaller than 5, use 4 
for the trial divisor. Since the first 5 figures of the dividend 
make a smaller number than the five figures of the divisor, use 
the first six figures of the dividend for the first trial dividend. 
Since 4 is contained in 15, 3 times, 3 is the first figure of the 
quotient. The second figure of the quotient is easily seen to 
be 5, and the second remainder is 3673; annexing 7, the next 
figure of the dividend, the third trial dividend, 36737, is smaller 
than the divisor; hence, write a cipher (0) for the third figure 
of the quotient, and annex 4, the next figure of the dividend, 
making the fourth trial dividend 367374. While 36 -^ 4 = 9, 9 
is evidently too large, since 43 X 9 = 387; consequently, try 8 
for the fourth figure of the quotient. The fifth figure is 4, and 
the remainder is 10758, which is written over the divisor, as shown. 
Applying the check, 1 + 5 + 2+1 + 8 + 7 + 4 + 6 = 34, 
and 3 + 4 = 7; 4 + 3 + 6 + 7 = 20, or 2; 3 + 5 + 8 + 4 = 
20, or 2; 1 + 7 + 5 + 8 = 21, and 2 + 1 = 3; then 2X2 = 4, 
and 4 + 3 = 7, the same remainder as was obtained for the divi- 
dend; hence, the work is probably correct. 

68. Rule I. — Write the divisor to the right of the dividend, with 
a line between, and draw a line under the divisor. 

II. Determine the trial divisor as previously described and divide 
it into the first trial dividend for the first figure of the quotient; multi- 
ply the divisor by this figure, subtract the product from the trial divi- 
dend, and annex the next figure of the dividend for a new trial dividend. 
Divide the second trial dividend by the trial divisor for the second 
figure of the quotient. Proceed in this manner until all the figures 
of the dividend have been used. 

III. If any trial dividend is smaller than the divisor, write a 
cipher for the corresponding figure of the quotient, and annex the 
next figure of the dividend for a new trial dividend. 

IV. If there is a remainder, write it over the divisor with a line 
between, and annex this expression to the quotient. 



EXAMPLES 

220^ 

(1) Divide 31415927 by 4726. Ans. 6647^^- 

7SQ0 

(2) 40073836 ^ 8018 = ? Ans. 4997gQjg- 

,„. 712946 ^ , ,o„o359 

(3) -^^g- = ? Ans. 13733^. 



34 ARITHMETIC §1 

(4) Divide 43560 by 209. Ans. 208209- 

(5) 30159681 ^ 5307 = ? Ans. 5683. 
4,396,652,679 _ ^ 51009^^^ 

^^^ 86193 ~ • ^''^- ^^""^86193- 

(7) Divide 2,189,404,900 by 29,950. Ans. 73,102. 

(8) According to Bessel, the diameter of the earth at the equator is 41,- 

847,192 feet; what is the diameter in miles, one mile containing 5280 feet? 

,,^^.3192 
Ans. 7925^2^ miles. 

(9) How many reams of 480 sheets each are contained in 75,960 sheets of 

120 
writing paper? Ans. ISSjg^ reams. 

(10) How many bales of rags can be made up from 238,996 pounds of 
rags, if the bales average 596 pounds each? Ans. 401 bales. 

(11) 12,656 pounds'of paper is to be put up in reams of 25 pounds each; 
how many reams will it make? Ans. 506 2 5 reams. 

(12) The freight bill on a shipment of pulpwood called for payment on 
246,782 pounds; if the average weight of a cord is 4450 pounds, how many 

rr2032 

cords are there? Ans. 55 . .-^ cords. 



SOME PROPERTIES OF NUMBERS 



DIVISIBILITY OF NUMBERS 

69. As previously stated, the factors of the product of two 
numbers are the two numbers which, when multiplied together 
produce the product. If more than two numbers are multiplied, 
the product has more than two factors; thus, 4 X 7 X 12 X 25 = 
8400, and 8400 may be considered to have as its factors 4, 7, 12, 
and 25. Since 12 = 3 X 4, and 25 = 5 X 5, 8400 also has as its 
factors 3, 4, 4, 5, 5, and 7, because 3X4X4X5X5X7 = 
8400. These factors may be combined in any way to form other 
factors; thus, 3 X 5 = 15, 4 X 5 = 20, and 4 X 15 X 20 X 7 = 
8400, or 4 X 5 X 5 = 100, and 3 X 4 X 7 X 100 = 8400, etc. 

70. A multiple of a number (the given number) is a certain 
number of times the given number; thus, 24 is a multiple of 6, 
6 being the given number, because 4 times 6 is 24 ; it is a multiple 
of 8, because 3 times 8 is 24; it is a multiple of 12, because 2 times 
12 is 24. In other words, any number that can be expressed 
as the product of two or more factors is a multiple of any one of the 



§1 SOME PROPERTIES OF NUMBERS 35 

factors or of the product of two or more of its factors. For in- 
stance, 7854 = 2 X 3 X 7 X 11 X 17; it is, therefore, a multi- 
ple of any one of these numbers and may be exactly divided by any 
one of them, and when it has been divided by one of them, the 
quotient may be exactly divided by any one of the remaining 
factors; thus, 7854 ^ 11 = 714, and 714 may be exactly divided 
by any of the remaining factors, since 714 -r- 2 = 357; 357 -r- 3 
= 119; 119 -^ 7 = 17, the last factor. A multiple of several fac- 
tors may be exactly divided by the product of any number of 
those factors; thus, 7854 h- 11 X 17 = 7854 -- 187 = 42; 7854 
-f- 3 X 7 X 11 = 7854 -=- 231 = 34; etc. 

When a number can be exactly divided by another number, the 
first number is said to be divisible by the second number; if, 
however, there is a remainder after the division, then the first 
number is not divisible by the second. For example, 84 is divis- 
ible by 2, 3, 4, 6, 7, 12, 14, 21, 28, and 42, and by no other num- 
bers except itself (84) and 1; 84 is also, of course, a multiple of 
these numbers. 

71. An odd number is one whose last (right-hand) figure is 1, 3, 
5, 7, or 9; 71, 423, 625, 1007, 1649 are odd numbers. 

An even number is one whose last figure is 0, 2, 4, 6, or 8; 640, 
972, 1774, 31416, and 2008 are even numbers. 

Any even number is divisible by 2; thig may be considered as 
another definition of an even number. Thus, any of the above 
even numbers are divisible by 2. 

Any number ending in 5 is divisible by 5; thus, 635, 895, etc. 
are divisible by 5. 

Any number ending in is divisible by 5 and by 10; thus, 640 
= 64 X 10 = 64 X 2 X 5. Since any number ending in is a 
multiple of 10, it has for two of its factors 2 and 5; it is therefore 
divisible by 2, 5, and 2 X 5 = 10. See Art. 70. 

72. A number that is not divisible by any number except itself 
and 1 is called a prime number or a prime; all prime numbers 
except 2 are odd numbers, since any even number is divisible by 
2. The prime numbers less than 100 are 1, 2, 3, 5, 7, 11, 13, 17, 
19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. 

A number that is divisible by some number other than itself 
and 1 is called a composite number. Composite numbers may be 
odd or even, and may always be expressed as the product of two 
or more prime numbers; thus, 84 = 2X2X3X7; 105 = 3X5 



36 ARITHMETIC §1 

X 7; 69 = 3 X 23; etc. When the factors are prime numbers, 
they are called prime factors. 

(a) Any number ending in two ciphers is divisible by 4 and by 
25; thus, 62100 = 621 X 100 = 621 X 4 X 25, since 100 = 4 
X 25. 

(6) Any number ending in three ciphers is divisible by 8 and by 
125; thus, 621000 = 621 X 1000 = 621 X 8 X 125, since 1000 
= 8 X 125. 

Since 100 = 2 X 50 = 20 X 5, a number ending in two 
more ciphers is divisible by 2, 4, 5, 10, 20, 25, and 50. 

Since 1000 = 2 X 500 = 4 X 250 = 5 X 200 = 8 X 125 = 
10 X 100 = 20 X 50 = 40 X 25, a number en ding in three or more 
ciphers is divisible by 2, 4, 5, 8, 10, 20, 40, 50, 100, 125, 200, 250, 
and 500. 

(c) If the sum of the digits of any number is a multiple of 3, 
the number is divisible by 3; when adding the digits, it is not 
necessary to add 3, 6, or 9. Thus, 31416 is divisible by 3, because 
1 + 4+1 = 6, a multiple of 3 ; similarly, 2350976 is not divisible 
by 3, because 2 + 5 + 7 = 14, which is not a multiple of 3. 

(d) If the number is even and the sum of the digits is a multi- 
ple of 3, the number is divisible by 6; thus, 31416 is divisible by 
6, because it is even and the sum of the digits is a multiple of 3. 
Similarly, 790108 is not divisible by 6, because 7 + 1 + 8 = 16, 
which is not a multiple of 3. Had the number been 780108, 
7 + 8 + 1 + 8 = 24, a multiple of 3, and the number is divisible 
by 6. No odd number is divisible by 6, 

(e) If the sum of the digits is a multiple of 9, the number is 
divisible by 9. Thus, 2,072,322 is divisible by 9, because 2+7 
+ 2 + 3 + 2 + 2 = 18, a multiple of 9; and 3,263,016 is not 
divisible by 9, because 3 + 2 + 6 + 3 + 1 + 6 = 21, which is 
not a multiple of 9; it is a multiple of 3, however, and since the 
number is even, it is divisible by 3 and by 6. 

(/) A number is divisible by 4 when the last two figuies are 
divisible by 4; thus, 4692, 543,836, 127,324, etc. are all divisible 
by 4, because 92, 36, and 24, the last two figures of each number 
are divisible by 4. But, 1742, 67,583, 98,782 , etc. are not divisible 
by 4. 

(g) A number is divisible by 8 when the last three figures are 
divisible by 8; 1752, 57064, 31416, etc. are divisible by 8, because 
752, 064, and 416, the last three figures of each number are divis- 
ible by 8. But 1852, 57164, and 31426 are not divisible by 8. 



§1 SOME PROPERTIES OF NUMBERS 37 

CANCELATION 

73. When one composite number is to be divided by another 

composite number or when the product of several numbers is to 

be divided by the product of several other numbers, the work 

can frequently be shortened by employing the process called 

cancelation. Cancelation means dividing out or canceling equal 

factors from the dividend and divisor. For example, suppose 216 

were to be divided by 36. According to Art. 72, 216 is divisible 

by 4 and by 9, and 36 is also divisible by 4 and by 9. Indicating 

216 
the division by -^, divide both dividend and divisor by 4, and 

the result is V; again dividing both dividend and divisor, this 
time by 9, the result is f = 6, and 6 is the quotient of 216 -i- 36. 

6 

In practice, the work would be performed as follows : -^ = 6. 

1 
Here 216 is divided by 4, a line is drawn through it, thus striking 
out or canceling 216, and the quotient, 54, is written above it; 
36 is also divided by 4, canceled, and 9, the quotient, is written 
below it. Since 54 and 9 have a common factor, 9, 54 is divided 
by 9, canceled, and the quotient, 6, is written above it ; 9 is divided 
by 9, canceled, and the quotient, 1, is written under it. Since 6 
4-1 = 6, the quotient of 216 -^ 36 is 6. The same result might 
have been obtained by resolving the dividend and divisor into 
their prime factors and canceling those that are common; thus, 

216 _ ^X^X2X^X^X3 _ 2 X 3 - 6 When the auo- 
W ^X2X^X^ -2X3- b. When the quo 

tient is 1, it is not customary to write it when canceling. 

Canceling the same factor in both dividend and divisor does not 
alter the value of the quotient. 

74. If there are several factors in both dividend and divisor, the 
process is similar; any factor common to both dividend and divisor 
may be canceled. For example, divide 136 X 54 X 117 by 26 X 51 
X 40. Writing the dividend over the divisor, with a line between, 

9 

;y ^/ 9 

M Xf>^Xm 9X9 81 



^^ X ^I X ^0 5 5 

;^ ;; 5 



16i 



38 ARITHMETIC §1 

By Art. 72, 136 is divisible by 8; 40 is also a multiple of 8; hence, 

cancel 136 and 40, writing the quotients as shown. Since 26 

and 54 are divisible by 2, cancel and write the quotients as shown. 

By Art. 72, 51 is divisible by 3, and since 27 is a multiple of 3, 

cancel and write the quotients as shown. Since there is a 17 

above the line and another below it, cancel the 17's. By Art 

72, 117 is divisible by 9; it is equal to 9 X 13; hence, divide the 

13 below the line into 117. As there are no more common factors, 

9X9 
the original expression is equal to — ^ — > being the product of all 

the uncanceled factors above the line divided by the product of 
all the uncanceled factors below the line. 

As another example, what is the value of 

224 X 5700 X 189 X 85 _ ^ 
250 X 456 X 21 X 17 ~ • 

Here, 

2 

^ 9 

28 ^70 ^f> in 
m X m^ X if>^ X ^^ _ 28 y 2 X 9 - 504 



X m X n xn 

By Art. 72, 224 and 456 are both divisible by 8; hence, cancel 
and write the quotients as shown. Cancel the 10 in 250 and 5700, 
leaving 25 and 570. Cancel the 57 in the divisor into 570 in the 
dividend. Cancel 5 in 25 and 10, leaving 5 and 2, and cancel the 
5 into 85, leaving 17, which cancels 17 in the divisor. 189 and 
21 are both divisible by 3, and 63 is a multiple of 7. Canceling 
as shown, all the factors in the divisor have been canceled, leaving 
28 X 2 X 9 = 504 for the quotient. That this is correct may be 
proved by actual multiplication and division ; thus 224 X 5700 X 
189 X 85 = 20,511,792,000; 250 X 456 X 21 X 17 = 40,698,- 
000; and 20,511,792,000 -^ 40,698,000 = 504. 

Rule. — Cancel the factors that are common to both dividend and 
divisor, and divide the ^product of all the factors that remain in the 
dividend by the product of all the factors that remain in the divisor. 



§1 SOME PROPERTIES OF NUMBERS 39 

EXAMPLES 

,,. 39 X 162 X 87 X 96 _ „ . _„3 

^^^ 19 X 6 X 29 X 42 X 13 " ' ^'''- ^^^• 

30 X 40 X 50 X 60 X 70 _ , 

^^^ 15 X 25 X 35 X 45 X 55 ~ ^ ^'''- ^^^• 
, . 24 X 44 X 64 X 84 X 104 ^ 
*• '^ 33 X 32 X 42 X 52 

... 231 X 328 X 7200 „ a n^ox 

(^) 21 X 64 X 5 25- = " ^"^- ^^^- 

31416 X 55 X 192 _ , 

^^^ 121 X 56 X 85 ~ • ^'''- ^^^• 

... 5236 X 630 X 128 X 192 „ . _„_, 

^^) 196 X 32 X 144 X 90 = ^ ^"^- ^^^*- 



= ? Ans. 256. 



AEITHMETIC 

(PART 1) 



EXAMINATION QUESTIONS 

(1) Express in figures the following numbers: (a) ten million 
nineteen thousand forty-two; (6) seventy thousand six hundred 
five; (c) five hundred sixty-two hundredths. 

(2) Express the following numbers in Roman notation: (a) 
4068; (b) 44; (c) 657,903; (d) 1920; (e) 1888. 

(3) What is the sum of $18.04, $1.57, $197.85, $36.43, and 
$360.52? Ans. $614.41. 

(4) Seven barrels, with their contents, weigh respectively 297 
pounds, 417 pounds, 226 pounds, 388 pounds, 293 pounds, 185 
pounds, and 313 pounds; if they are all hoisted in an elevator at 
one time, and the elevator weighs 2309 pounds, together with two 
men, one weighing 148 pounds and the other 187 pounds, what is 
the total weight lifted? Ans. 4763 pounds. 

(5) From a pile of pulp weighing 12,882 tons, 2057 tons were 
removed on a certain day and 3836 tons on another day; how 
many tons remained in the pile? Ans. 6989 tons. 

(6) If 256,405 be subtracted from a certain number and the 
remainder is 700,999; (a) what is the number? (6) What num- 
ber must be subtracted from 1,001,010 to give a remainder of 
660,019? ^^ /(a) 957,404 



(6) 339,991. 

(7) What is the product of 461, 217 and 865. Check the re- 
sult by casting out 9's. Ans. 86,532,005. 

(8) If the quotient is 5077, the divisor is 6345, and the re- 
mainder is 2609, what is the dividend? Check by casting out 
9's. Ans. 32,216,174. 

(9) If the dividend is 4,511,856, the quotient is 2803, and the 
remainder is 1829, what is the divisor? Ans. 1609. 

(10) There are 5280 feet in a mile; (a) how many miles in 
408,806 feet? (6) How many feet in 1894 miles? 

Ans / ^^) ^^^*^^ '^'^^^' 
' 1(6) 10,000,320 feet. 

41 



42 ARITHMETIC §1 

(11) There are five numbers: (1) 840, (2) 231, (3) 1728, (4) 
2618, and (5) 1215; (a) which of these numbers are odd? (h) 
which are even? (c) which are divisible by 3? (d) by 6; (e) by 9; 
(/) by 12? 

(12) Referring to the last question, which of the numbers are 
divisible (a) by 2? (6) by 4? (c) by 8? (d) by 5? 

/-,o^^.•^.l, 1 , 5236 X 9 X 45 X 26 ^ . ^. 

(13) Fmd the value of — qi y iqt^ V 88 — '^ cancelation. 

Ans. 51. 

(14) A hogshead contains 63 gallons; how many hogsheads are 
equal to 20,610 gallons? Ans. 327/-3 hogsheads. 

^.^^T.■A.^. 1 .1309X1728X448, _. 

(15) Fmd the value oi o^fi V 88 V 1^4 -^ cancelation. 

Ans. 291tt. 



ARITHMETIC 

(PART 2) 



GREATEST COMMON DIVISOR AND LEAST 
COMMON MULTIPLE 

75. Greatest Common Divisor. — The greatest common divisor 

of two or more numbers is the largest number that will exactly 
divide all of them. Thus, the greatest common divisor (usually- 
abbreviated G.C.D.) of 36 and 48 is 12, because 12 is the largest 
number that will divide 36 and 48 — it is the largest common 
factor of 36 and 48; also, 36 is the G.C.D. of 72, 108, and 144, 
because it is the largest common factor of those numbers. 

To find the G.C.D. of two numbers, find the product of their 
common factors; the work is usually arranged as follows : Find the 

G.C.D. of 54 and 216. Write the 
' numbers as shown in the margin, 



' separating them by a comma, and 



G C D =6 X 9 = 54 Ans ^^^'^ proceed as in short division. 

'By Art. 72, 54 and 216 are both di- 
visible by 9; divide as shown, obtaining 6 and 24 for the quotients. 
Both 6 and 24 are divisible by 6, the quotients being 1 and 4. 
The only factor common to 1 and 4 is 1, which has no effect on 
the G.C.D. Since 54 and 216 have the common factors 9 and 6, 
they are divisible by the product of these common factors, and 
the G.C.D. of 54 and 216 is 9 X 6 = 54. To prove the work, 
54 ^ 54 = 1, and 216 4- 54 = 4. 

76. To find the G.C.D. of more than two numbers, proceed in 
exactly the same way as in Art. 75, remembering that the divisors 
must divide all the numbers. Thus, find the G.C.D. of 270, 

405, 675, and 945. Arrang- 
ing the work as shown, it is 
seen that all the numbers are 
multiples of 9; hence, divide 
by 9. The quotients are all 
divisible by 5; hence, divide 
43 





9 
5 
3 


270, 405, 


675, 945 




30, 45, 


75, 105 




6, 9, 


15, 21 


G.C.D. 


= 9 


2, 3, 
X5 X3 


5, 7 
= 135. Ans 



44 



ARITHMETIC 



391 


238 


1 


238 


153 


1 


153 


85 


1 


85 


68 


1 


68 


17 


4 


68 













G.C. 


D. = 


17. 



Ans. 



by 5. The quotients in the third line are all multiples of 3; 
hence, divide by 3. The quotients in the fourth line have no 
common factor; therefore, the G.C.D. is 9X5X3 = 135. 
To prove this, divide each of the original numbers by 135; the 
quotients will be the same as in the bottom line. 

77. It frequently happens that the common factors are not 
readily seen; in such cases, proceed as in the following example. 
Find the G.C.D. of 238 and 391. Write the larger number, and 

draw a vertical line on its right; then 
write the smaller number and draw 
another vertical line. The first vertical 
line is the sign of division, but the 
second vertical line is merely a line of 
separation. Now divide 391 by 238, 
and write the quotient to the right of 
the second vertical line. The remainder 
is 153, which is divided into 238, the quotient being written under 
the first quotient. The remainder from the second division is 
85, which is divided into 153, giving a remainder of 68, the quo- 
tient being written under the preceding quotient. Dividing 85 
by 68, the remainder is 17. Lastly, 68 divided by 17 gives a 
quotient of 4 and a remainder 0; and 17, the divisor that gave the 
remainder 0, is the G.C.D. of 238 and 391. 

As another example, find the G.C.D. of 31,416 and 1,400,256. 
Arranging the work as before, the first 
quotient is 44, and the remainder is 
17,952, which is divided into 31,416. 
The second quotient is 1, and the second 
remainder is 13,464, which is divided 
into 17,952. The third quotient is 1, 
and the third remainder is 4,488, which 
is contained in 13,464 3 times, and there 
Therefore, the G.C.D. of 31,416 and 1,400,256 



1400256 

125664 

143616 

125664 

17952 

13464 

4488 


31416 
17952 

13464 
13464 





44 
1 
1 
3 



G.C.D. = 4488. Ans. 



is no remainder 
is 4488. 

It is seldom necessary to find the G.C.D. of more than two 
numbers. The first method should be used whenever possible; 
but, when the numbers are large or the common factors are not 
apparent, then use the second method. Observe that the process 
must be carried out until a remainder of is obtained. If the 
numbers have no common divisor, the G.C.D. will be 1. In 
such case, the numbers are said to be prime to each other, though 



§1 LEAST COMMON MULTIPLE 45 

both may be composite numbers. For example, 91 and 136 are 

both composite, but they are prime to 

each other, as is evident when the 

greatest common divisor is found to be 

1, in accordance with the work as shown 

^ ^ -P, " . in the margin. 

G.C.D. = 1. Ans. ° 



136 


91 


1 


91 


90 


2 


45 


1 


45 


45 













= 1. 


Ans. 





EXAMPLES 

(1) Find the G.C.D. of 40, 60, 80, and 100. Ans. 20. 

(2) Find the G.C.D. of 70, 175, 210, and 245. Ans. 35. 

(3) Find the G.C.D. of 2387 and 1519. Ans. 217. 

(4) Find the G.C.D. of 4059 and 6390. Ans. 9. 

(5) Find the G.C.D. of 11433 and 444. . Ans. 111. 
(.6) Find the G.C.D. of 364089 and 457368. Ans. 3009. 
(7) Find the G.C.D. of 1016752 and 991408. Ans. 176. 



78. Least Common Multiple. ^ — ^A common multiple of several 
numbers is a number that is divisible by those numbers; thus, 
840 is a common multiple of 40 and 56. 

The least common multiple of two or more numbers is the 
smallest number that is divisible by those numbers; thus, 280 is 
the least common multiple of 40 and 56 ; it is the smallest number 
that is divisible by both 40 and 56. 

The product of two or more numbers is a common multiple of 
the numbers; for the want of a special name, this will be called 
the prime multiple of the numbers. The prime multiple may be 
the least common multiple, and will be such if no two of the 
numbers have a common factor; thus, the prime multiple of 91 
and 36 is 91 X 36 = 3276, and this is also the least common 
multiple of 91 and 36. The words least common multiple are 
usually abbreviated to L.C.M. If the two numbers have a com- 
mon factor, it may be canceled from the numbers, and the L.C.M. 
will then be the product of this common factor and the remaining 
factors of the numbers. Thus, 56 and 72 have the common factor 
8; dividing 56 and 72 by 8, the quotients are 7 and 9, respectively; 
then the L.C.M. is 8X7X9 = 56 X9 = 7X8X9 = 7X 
72 = 504. Note that 504 is equal to 56 X 9 and to 72 X 7; 
hence, it is divisible by both 56 and 72. Since 8, the common fac- 
tor, is the greatest common factor, that is, the G.C.D., of 56 and 
72, it follows that the L.C.M. of two numbers may be found by 



46 



ARITHMETIC 



§1 



dividing one of the numbers by their G.C.D. and multiplying the 
other number by the quotient. For instance, referring to ex- 
ample (3) at the end of Art. 77, the G.C.D. of 2387 and 1519 is 
217; 1519 ^ 217 = 7; then 2387 X 7 = 16,709, the L.C.M. of 
1519 and 2387. Or, 2387 ^ 217 = 11, and 1519 X 11 = 16,709. 
79. To find the L.C.M. of more than two numbers, proceed in 
much the same way as in finding the G.C.D. An example will 
illustrate the process. Find the L.C.M. of 16, 24, 30, and 32. 
Arrange the work as shown. Since all the numbers contain the 
common factor 2, divide it out. In the first line of quotients, 



2 


16, 


24, 


30, 


32 


4 


8, 


12, 


15, 


16 


2 


2, 


3, 


15, 


4 


3 


1, 


3, 


15, 


2 




1, 


1, 


5, 


2 



L.C.M. =2X4X2X3X5X2= 480. Ans. 
three of the numbers are multiples of 4(= 2 X 2), but the other 
number contains no factor in common with 4; hence, divide by 4. 
The second line of quotients contains two numbers that are 
multiples of 2; hence, divide by 2. The third line of quotients 
contains two numbers that are multiples of 3; hence, divide by 3. 
The numbers in the fourth line of quotients are all prime to each 
other, and no further division is possible. The product of the 
divisors and the factors in the last line of quotients is the L.C.M., 
which is equal to 2X4X2X3X5X2 = 480. 

When finding the L.C.M., divide out all factors that are com- 
mon to two or more numbers until a row of quotients is obtained 
that are prime to one another. As another example, find the 
L.C.M. of 15, 27, 55, and 99. There is no factor common to all 



5 


15, 27, 55, 99 


11 


3, 27, 11, 99 


3 


3, 27, 1, 9 


3 


1, 9, 1, 3 




1, 3, 1, 1 



L.C.M. =5X11X3X3X3 = 1485. Ans. 
the numbers; but, since 15 and 55 are multiples of 5, divide by 5. 
Since 11 and 99 are multiples of 11, divide by 11. The remainder 
of the work is evident. Note that whenever a number is not 
divisible by a divisor, the number is brought down into the l^ne 
of quotients. The L.C.M. is5XllX3X3X3 = 1485. 



§1 



LEAST COMMON MULTIPLE 



47 



If the numbers are such that their factors are not apparent, find 
the L.C.M. of two of them, and then find the L.C.M. of this result 
and the third number; then the L.C.M. of the second result and 
the fourth number, and so on; the last result will be the L.C.M. of 
all the numbers. For instance, to find the L.C.M. of 893, 1387, 
and 1121, first find the L.C.M. of 893 and 1387 (or 1121). Pro- 
ceeding as described in Art. 78, first find the 
G.C.D. of 893 and 1387. The work is 
shown in the margin, and the G.C.D. is 19. 
Then, 893 -^ 19 = 47, and 1387 X 47 = 
65189, the L.C.M. of 893 and 1387. Now 
find in the same way the L.C.M, of 1121 
and 65,189. The work for this is also 
shown in the margin. The G.C.D. of these two numbers is 19; 
1121 -^ 19 = 59; and 65189 X 59 = 3,846,141, which is the 
L.C.M. of 893, 1387, and 1121. 

It may here be remarked that 
the greatest common divisor and 
the least common multiple are of 
importance in connection with 
the reduction, addition, and 
subtraction of fractions, as will 
shortly appear. 



1387 


893 


893 


494 


494 


399 


399 


380 


95 


19 


95 










65189 

5605 

9139 

8968 

171 

95 

76 

76 



1121 -- 19 = 59 
65189 X 59 = 3846151. A 



1121 

1026 

95 

76 

19 



58 
6 
1 
1 
4 



(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 



Find the L.C.M. 
Find the L.C.M. 
Find the L.C.M. 
Find the L.C.M. 
Find the L.C.M. 
Find the L.C.M. 
Find the L.C.M. 



EXAMPLES 

of 28, 49, 63, and 84. 
of 12, 14, 16, and 18. 
of 15, 20, 25, and 30. 
of 4, 8, 12, 16, and 20. 
of 1955 and 4403. 
of 119, 204, 272. 
of 442, 234, and 1001. 



Ans. 1764. 

Ans. 1008. 

Ans. 300. 

Ans. 240. 

Ans. 506,345. 

Ans. 5712. 

Ans. 306,306. 



48 ARITHMETIC §1 

FRACTIONS 



DEFINITIONS 

80. When an integer or a unit is divided into equal parts, one or 
more of these parts is called a fraction of the integer or unit. If, 
for example, a straight stick be cut into two pieces of the same 
length, one piece is equal to the other, and either is called one- 
half of the stick. If the stick is cut into three equal pieces or 
parts, one of them is called one-third of the stick; if cut into four 
equal parts, one of them is called one-fourth of the stick; if cut into 
five equal parts, one of them is called one-jifth of the stick; etc. 
The expressions one-half, one-third, one-fourth, etc. are fractions. 
More than one part is denoted by writing the number before the 
name of the part; thus, two-thirds means two one-thirds, three- 
fourths means three one-fourths, etc. 

81. To express a fraction with figures, it is necessary to write 
two numbers, one to show into how many parts the integer or 
unit has been divided, and the other to show how many of these 
parts are taken or considered. For instance, 

I means one-half, and indicates one of two equal parts 
I means one-third, and indicates one of three equal parts 
I means one-fourth, and indicates one of four equal parts 
f means two-thirds, and indicates two of three equal parts 
-f means four-fifths, and indicates four of five equal parts 
tV means seven-twelfths, and indicates seven of twelve equal 
parts, etc., etc. 

82. It will be observed that a fraction is expressed in one of the 
ways used to indicate division. The number below the line is 
called the denominator, because it denominates, or names, the 
number of parts into which the integer has been divided. 

The number above the line is called the numerator, because it 
numerates, or counts, the number of the equal parts that are taken 
or considered. 

83. An expression like f may be interpreted in two ways: 
First, it is 3 times i of a unit. For example, one dollar is equal 
to 100 cents; one-fourth of a dollar is 100 ^ 4 = 25 cents; and 
three-fourths of a dollar is 3 X 25 cents = 75 cents. Here one 
dollar is the unit. Second, it is one-fourth of three times the unit. 
If the unit is one dollar, three times the unit is 3 dollars or 300 
cents, and one-fourth of 3 times the unit is 300 cents -^ 4 = 75 



§1 FRACTIONS 49 

cents, the same result as before. In the first case, f is a fraction; 
in the second case, it is an indication of division. 

84. Except when the denominator is 1, 2 or 3, a fraction is 
read by pronouncing the name of the numerator and then pro- 
nouncing the name of the denominator after adding ths; thus, f 
is read three-sevenths, tI is read eleven-sixteenths, tit is read 
forty-nine one-hundred-twenty-fifths, etc. But if is read thirteen- 
twenty-firsts; |f is read twenty-nine-forty-seconds; || is read 
thirty-seven fifty-thirds, etc. 

85. If two fractions have the same numerator, but a different 
denominator, the fraction whose denominator is the smaller is 
the larger; thus f is larger than f , because one-fifth of anything 
is smaller than one-fourth of it; hence, 3 one-fifths is smaller than 
3 one-fourths For example, three-fourths of a dollar is 75 cents, 
but three-fifths of a dollar is 60 cents, since one-fifth of a dollar 
is 100 cents ^ 5 = 20 cents, and 3 X 20 cents = 60 cents. 

86. The numerator of a fraction may be greater than the de- 
nominator, in which case, the value of the fraction is found by 
dividing the numerator by the denominator; thus, V" = 4, the 
value of the fraction; ^- = 2, the value of the fraction; etc. If 
the numerator is equal to the denominator, the value of the fraction 
is 1; thus, TT = 1, I = 1, etc. This is evident, since if a dollar 
is divided into, say, 5 equal parts, 5 of these parts make up the 
dollar. If the numerator is less than the denominator, the 
value of the fraction is less than 1. 

87. A proper fraction is one whose numerator is smaller than 
its denominator; its value is always less than 1. Thus, f, tV, 
tI , etc. are proper fractions. 

An improper fraction is one whose numerator is equal to or 
greater than its denominator. Thus, f, 1, ||, etc. are impro- 
per fractions. 

When it is not desired to specify the numerator and denomina- 
tor separately, they are called the terms of the fraction; thus, the 
terms of the fraction |f are 23 and 42. 

When a fraction is joined to an integer, as in the expression 
14|, the expression is called a mixed number. Here 14f means 
14 and f more; it has the same meaning as 14 -f f . In reading 
a mixed number, the word and is used as above, but the plus sign 
is always understood, though not written or spoken; thus, 5f is 
read five and two-thirds, and means 5 + f . Mixed numbers 



50 ARITHMETIC §1 

always occur whenever the dividend is not a multiple of the divi- 
sor; for instance, the quotients found in examples (3), (4), (6), and 
(7) of Art. 65 are mixed numbers. 

88. In printing and writing, in order to save space, fractions are 
frequently expressed by using the inclined line instead of the 
horizontal line; thus, M = f > 113/147 = lif, etc. In such 
cases, a hyphen is sometimes written between the integer and the 
fraction of a mixed number; thus, 14-3/8 = 14%; the hyphen 
shows that the fraction belongs to the integer. 

When it is desired to indicate that the fraction is to be pro- 
nounced and it is not desired to write the name in full* the fraction 
is written as usual and the ending of the name of the denominator 
is annexed; thus, ^^rds, %2ths, ^%2ds, ^^j^sts, mean two-thirds, 
seven-twelfths, fifteen twenty-seconds, twenty-three thirty-firsts, 
etc. The only exception is ^, which is always read one-half, and 
is always so written and printed. 



REDUCTION OF FRACTIONS 

89. To reduce a fraction is to change its form without changing 
its value; to change its form means to alter its numerator and de- 
nominator. It was shown in Arts. 82 and 83 that a fraction may 
be regarded as an expression of division, the numerator being the 
dividend and the denominator the divisor. In Art. 73, it was 
shown that canceling the same factor in both dividend and divisor 
does not alter the value of the quotient; hence, dividing both 
numerator and denominator of a fraction by the same number 
does not alter the value of the fraction. For example, ^f = 
yV = I- Here both 18 and 24 are first divided by 2, the quo- 
tients being 9 and 12, respectively. Since 9 and 12 contain the 
common factor 3, divide both 9 and 12 by 3, the quotients being 
3 and 4, respectively. The numerator and denominator of the 
given fraction might both have been divided by their greatest 
common divisor, 6, and the same final result, f, obtained. 

That Hths = fths is easily shown. Thus, a gross is 144; 
i-|ths of a gross is 18 times ^th of 144; ;2Vth of 144 is 144 -f- 24 
= 6, and |-|ths is 18 X 6 = 108. But fths of a gross is 3 times 
ith of 144; |th of 144 = 36, and fths of 144 is 3 X 36 = 108. In 
the same way, it is shown that rVths of 144 is 108. Therefore, 

1 8 _ 9 _ 3 
"5T — TY — 4" 



§1 FRACTIONS 51 

90. Multiplying both numerator and denominator by the same 
number does not alter (change) the value of the fraction; this is 
evident, since both terms of the new fraction may be divided by 
the number used as a multiplier, thus obtaining the original frac- 

3X6 
tion. For instance, , ^ = |-|, which, as has just been shown 

2x7 
is equal to |. Similarly, - - = |-f- 

91. When the fraction is reduced to lower terms, that is, when 
the numerator and denominator are made smaller by division, 
the process is called reduction descending. When the fraction is 
reduced to higher terms, that is, when the numerator and de- 
nominator are made larger by multiplication, the process is called 
reduction ascending. 

92. When the fraction has been reduced by division until the 
numerator and denominator are prime to each other (have no 
common factor), the fraction is said to be in its lowest terms; 
thus, f, tV, if? etc. are fractions in their lowest terms. When 
a fraction in its lowest terms, it is in its simplest form. 

93. To reduce a fraction to its lowest terms, cancel all factors 
that are common to the numerator and denominator. If the 
factors are not apparent, and it is desired to make sure that the 
fraction is in its lowest terms, find the greatest common divisor, 
if any, of the numerator and denominator and divide both terms 
by it. 

Example 1. — Reduce ||| to its lowest terms. 

Solution. — Since both terms are multiples of 4, divide them by 4, and 
tI"! ~ ii- Both terms of the new fraction are multiples of 7; hence, divid- 
ing by 7, If = f. Both terms of the new fraction are multiples of 3; 
hence, dividing by 3, f = f. Since the terms are now prime to each 
other, the fraction is in its lowest terms. In practice, the work would be 
arranged as follows : -^-f -f =11=1 = 3. Ans. 

Example 2. — Reduce f Hit to its simplest form. 

Solution. — Since both terms are multiples of 8 (see Art. 72), divide them 
by 8, and f gff f = f f -ff. Apparently, the terms have no common factor, 
but to make certain, apply the process of finding the G.C.D. of 6409 and 
8671; in this case, the G.C.D. is 377. Dividing both terms by 377, ||ff 

94. To reduce a fraction to another fraction having a given 
denominator, divide the given denominator by the denominator 
of the given fraction and multiply the terms of the fraction by 
the quotient. For example, to reduce f to a fraction having 96 



52 ARITHMETIC §1 

3 X 12 36 
for its denominator; 96 -^ 8 == 12, and o y io = qa- Arts. 

The reason for dividing the given denominator by the denomina- 
tor of the given fraction is evident, since the quotient must be the 
number by which the denominator of the given fraction must be 
multipHed in order to equal the given denominator. This opera- 
tion is of importance in adding and subtracting fractions. 

95. To reduce an integer to an improper fraction having a given 
denominator, multiply the integer by the given denominator, and 
the product will be the numerator of a fraction having the given 
denominator. For instance, reduce 7 to a fraction having 12 for 
its denominator. Here 7 X 12 = 84, and the required fraction 
is ff . That this result is correct may be proved by dividing the 
numerator by the denominator, the quotient being 7, the original 
number. This result may also be obtained in another way; thus, 

7 is evidently equal to t, and r -lo — Jo- 

96. To reduce a mixed number to an improper fraction, multi- 
ply the integral part by the denominator of the fraction, add the 
numerator to the product, and write the sum over the denomina- 
tor. This is evidently correct, since the mixed number is obtained 
by dividing the dividend (numerator) by the divisor (denomina- 
tor), and the remainder is written over the divisor. Thus, reduce 
14f to an improper fraction. Here 14 X 8 = 112; 112 + 3 
= 115; hence, 14| = ^-^. Regarding ^^ ths as an indication of 
dividing 115 by 8, 115 ^ 8 = 14f. 

97. A common denominator of two or more fractions is a com- 
mon multiple of the denominators of the fractions; and the least 
common denominator is the least common multiple of the denomi- 
nators. For instance, the least common denominator of f , |, 
and I is 24, because 24 is the L.C.M. of 3, 4, and 8. 

98. To reduce two or more fractions to fractions having a least 
common denominator, find the L.C.M. of the denominators; then, 
by the method of Art. 94, reduce each fraction to a fraction hav- 
ing this denominator. 

Example 1. — Reduce f, j, and f to fractions having a least common 
denominator. 

Solution.— The L.C.M. of 3, 4, and 8 is 24; f = if ; i = ^\; and f 

Example 2. — Reduce ||, f|, and j-f-f to fractions haying a least common 
denominator. 



§1 FRACTIONS 53 

Solution.— The L.C.M. of 69 and 92 is 276; the L.C.M. of 276 and 161 is 
1932, which is therefore the L.C.M. of 69, 92, and 161. Then, 1932 -- 69 
= 28, and || X || = ifff; 1932 - 92 = 21, and || X |i = HMj 1932 
-i- 161 = 12, and 111 X il = ilM- Therefore, the required fractions are 

1232 1197 „riH 1^4 0. 4„„ 

Example 3. — Which fraction is the larger, jW or Hg ? 

Solution. — To determine which is the larger, reduce them to a common 
denominator; then the one that has the greater numerator is the larger. 
Since 4f § = ||; reduce i%% and |f to a common denominator, preferably, 
the least common denominator. Since 133 and 39 have no common factor, 
their L.C.M. is their product, or 133 X 39 = 5187; then y^s = If if' and 
|5 = I Iff. Therefore, xf§ is the larger fraction. .4ns. 

Example 4. — There are 1760 yards in a mile; what fraction of a mile is 
550 yards? 

Solution. — Since there are 1760 yards in one mile, the number of miles 
or parts of a mile in 550 yards is 550 -r- 1760 = tV^tj = tVc" = A- 
Therefore, 550 yards is j^ths of a mile. Ans. 

Example 5. — Which is the greater 500 yards or -gV^ths of a mile? 

Solution. — Since there are 1760 yards in a mile, 500 yards = iWo 
= y^y**^ - ^%8 mile; -gV^ = if mile. Reducing these fractions to a com- 
mon denominator, the L.C.M. of 88 and 53 is 88 X 53 = 4664; U 
= \Ut and II = iff|. Therefore, 500 yards is a little greater than 
■^''^^^ths of a mile. Ans. 



EXAMPLES 

(1) Reduce to its lowest terms J|-|. Ans. j|. 

(2) Reduce to its lowest terms ||-J. Ans. |i. 

(3) Reduce to its lowest terms i|||. Ans. i*^. 

(4) Express 11 as a fraction having 24 for its denominator. Ans. W*-. 

(5) Reduce 125.{6 to an improper fraction. Ans. -W- 

(6) Reduce 852iH to an improper fraction. Ans. ^-^V/^- 

(7) WhichisthelargeryVs oriff? Ans. ^U- 

(8) Which is the larger |f or |f ? Ans. ff. 

(9) Which is the larger ^i or if? Ans. if. 

(10) Reduce to their least common denominator |, |, /g, and i\. 

AriQ J8 40 SO SI 
/I 'to. 4jj, 48, 5 J, -f-g. 

(11) Reduce to theirleast common denominator |,i, |, H, and H. 

Amo 2 53 210 lAO 4 95 45.5 

ii.nb. ^30) eaiJ' tisd) sso) 'SsiT' 

(12) A ream of writing paper contains 480 sheets; 328 sheets is what 
fraction of a ream? Ans. tJth ream. 

(,13) How many sheets are equal to T^ths of a ream of writing paper? 

Ans. 280 sheets. 

(14) Which is the larger, 200 sheets of writing paper or Hth ream? 

Ans. Hth ream. 

(15) How many sheets difference are there between 400 sheets of writing 
paper and fjds ream? Ans. 5 sheets. 

(16) If a car of sulphur weighs 40,000 pounds, what fraction of a car-load 
is 26,800 pounds? Ans. iV^th car-load. 



54 ARITHMETIC §1 

ADDITION OF FRACTIONS 

99. The sum of A and yV is rf , because 5 one-sixteenths 
and 9 one-sixteenths = 14 one-sixteenths = yfths. Like num- 
bers (see Art. 8) can be added, but unhke numbers cannot be 
added; thus, 4 feet cannot be added to 6 inches, but 4 feet can be 
added to 6 feet and 4 inches can be added to 6 inches. In an 
expression like yV + tV, the denominators show what is to be 
added, in this case 16ths, and the numerators show how many of 
the things indicated by the denominator are to be added ; it is 
exactly the same operation as adding 5 dollars and 9 dollars and 
obtaining 14 dollars for the sum. In one case, 16ths are added, 
and in the other case dollars are added. Consequently, if the 
denominators are alike, the sum of several fractions may be found 
by adding the numerators and writing the sum over the denomi- 

5 + 9 + 13 
nator. For instance, A + iV + tI = 7g = f ^ = ill, 

the last result being obtained by dividing the numerator by the 
denominator. 



100. If the denominators are unlike, the fractions cannot be 
added until they have been reduced to a common denominator, 
preferably, the least common denominator; thus, f cannot be 
added to y, because 2 one-thirds + 4 one-fifths has no particular 
meaning as it stands — the fractions have no common unit. But, 
f = H, i = H, and H + H = fl = 1 A. That this result is 
correct is readily seen. Thus, a bushel of wheat weighs 60 
pounds; hence, fds of a bushel weighs 40 pounds; fths of a 
bushel weighs 48 pounds; and the combined weight is 40 + 48 
= 88 pounds. Now yVth of 60 pounds is 4 pounds, and 
ff ths of 60 pounds is 22 X 4 = 88 pounds, the same result 
as before. 

Similarly, to add 4 feet and 6 inches, it is necessary to express 
the feet in inches or the inches in feet before adding. Since there 
are 12 inches in 1 foot, there are 12 X 4 = 48 inches in 4 feet; and 
48 inches + 6 inches = 54 inches. If it is desired to express the 
sum in feet, then it is necessary to divide 54 by 12, obtaining 
4tV = 4| feet. The same result might have been obtained by 
dividing the number of inches, 6, by 12, thus obtaining a frac- 
tion of a foot, which can then be added to the number of feet; 
thus, 6 -T- 12 = tV = h that is, one-half of a foot, and 4 feet 
+ I foot = 4 + I = 4| feet. 



§1 FRACTIONS 55 

101. Rule. — I. If the fractions have a common denominator, add 
the numerators and write the sum over the denominator. 

II. If the fractions do not have a common denominator, reduce 
them to fractions having a common denominator, -preferably, their 
least common denominator, and then add. 

III. If the sum is an improper fraction, reduce it to an integer or 
mixed number by dividing the numerator by the denominator. 

IV. // the sum is a proper fraction or a mixed number containing 
a proper fraction, reduce the fraction to its lowest terms. 

Example.— Add ^V, H, U, and |f. 

Solution. — Here the least common denominator is 240. Then, /^ + {^ 
^■iz^is _j_4o ,_i65 ^204 I 130- 140 + 165+204+130 639 _,,., 

-t- 2 -|- 2¥ - 24 + 810 -t- 210 "t" 2?(J 240 "^ 240 ^ ^^^ 

In practice, the work would be arranged as follows : 
7+11:17,18 140 + 165 + 204+130 ^53 ^ 

Here the denominator of the sum is written only once, the 
numerators of the fractions being written above it and added. 
It is seen at a glance that the sum of the numerators is greater 
than the denominator; hence, they are added separately and the 
improper fraction reduced to a mixed number, with the fractional 
part in its lowest terms. 

102. The Sum of Two Fractions.— When it is desired to add 
two fractions whose denominators are unlike and have no common 
factor, that is, when these are prime to each other, proceed as 
follows: Add yi and ^. Here the denominators have no com- 
mon factor; hence, the least common denominator is the pro- 
duct of the denominators. This product divided by either 
denominator gives the other denominator, which is used as a 
multiplier for the numerator. Therefore, multiply the denomina- 
tors for a new denominator; multiply the numerator of the first 
fraction by the denominator of the second, and the numerator 
of the second fraction by the denominator of the first and add 
theproducts, Thus,-H+ .1,11X^ + 19X13 ^264 + 247 

' ^ ^ 13 X 24 312 

— ^^ — 11.9 9. A 

Even if the denominators have a common factor, this method of 
adding two fractions is generally to be preferred, because it is 
quicker than reducing the fractions to their least common 
denominator. 



56 ARITHMETIC §1 

103. To add mixed numbers, add the integral and fractional 
parts separately; if the sum of the fractions is an improper frac- 
tion, reduce it to a mixed number and add to the sum of the 
integers. 

Example.— What is the sum of 23f, SlH, 28M, and 25||? 

Solution. — Arrange the numbers in the same manner as for addition 

233 _ 24 of integers. The least common de- 

3111 = If nominator of the fractions is evidently 

28|i = 11 64; hence, reduce the fractions to frac- 

^^ II tions havmg 64 for a denommator, 

lUy^ J 6 J" — ■^s¥ writing them as shown. Adding the 

numerators, the sum is 149, and the sum of the fractions is ^-^ = 2|i. 
Carry the 2 into the column of integers, which add, obtaining 109, to which 
annex the fraction, obtaining 109U for the sum of the mixed numbers. Ans. 

The sum might have been obtained by reducing the mixed 
numbers to improper fractions and then adding by the regular 
rule; but this takes longer and there is greater liability of making 
a mistake. 
As another example, add df, 5|, 7ri, and 6|. The least 
common denominator is 120, Reducing 
the fractions to fractions having 120 for a 
denominator, the sum of the numerators 
24H Ans. is 27.3, which divided by 120 gives a quo- 

tient of 2^-0^. Carrying the 2 to the sum of the integers and 
annexing the fraction, the sum of the mixed numbers is 24-f-g-; 
thus, 2 + 6 -F 7 + 5 + 4 = 24, and 24 -F ii = 24^^. 



H 




_*8 


' 


1^6 


6i 


= 


40 


7H 


= 


m 


6^- 




75 




l^TJ 



EXAMPLES 

(1) Add f, ^j,, A. Ans. IfH. 

(2) Add H and xV- Ans. lr%. 

(3) Add I, h I I, jS. Ans. lU- 

(4) Add III and fff . Ans. 1A¥/y. 

(5) Add 11 A, 8f, 9if, lOi Ans. 39fi. 

(6) Add 127f, 85f, 109^, 96|. Ans. 419tVt7. 

(7) Three kinds of dyes were added to a beater of paper stock, and 
weighed 9^'i ounces, 8% ounces and 13% ounces; how much dye stuff was 
used? Ans. 32)^4 ounces. 



SUBTRACTION OF FRACTIONS 

104. Since subtraction is the reverse of addition, fractions must 
be reduced to a common denominator before the subtraction can 

6 — 5 
be performed. For example, f — f = — ^ — = I- Here the 



§1 FRACTIONS 57 

common denominator is found as in addition, then the difference 
of the numerators is taken and the remainder is written over the 
common denominator. 

Example. — Find the difference between y^f and ^il"- 

Solution. — As it is not evident which fraction is the larger, assume that 
the first fraction is the larger ; if, later, it is found not to be such, the subtra- 
hend and minuend in the numerator can be transposed. Then, proceeding 
as in addition of two fractions, 
1 fi 17 106 X 268 - 217 X 133 28408 - 28861 _ . 

m - Ui = 133 X 268 = 58156— • ^^'' expression 

„ 1 „ . . . 28861 - 28408 
shows that the fraction -f^ -J is the larger, and the difference is ^eT^R 

105. To subtract one mixed number from another, subtract 
the integral and fractional parts separately. For example, sub- 
tract 26f from 42f . The least common de- 

42f = if nominator is 24; hence, reduce the fractions 

26§ = A to fractions having 24 for a denominator and 

16^^ ^\ subtract as shown. The difference is found 

to be 16^V- Ans. 
Example.— From 109t\ take 96|. 

Solution. — Here the fraction in the subtrahend is larger than the fraction 

lOOj^^ = if lit = ii in the minuend; therefore, proceed exactly 

961^ = If ft as in subtraction of integers when the figure 

in the subtrahend denotes a larger number 

12||. Ans. If than the figure over it in the minuend, by- 

adding 1 to the fraction in the minuend and 1 to the integer in the subtrahend 
Reducing the mixed number, l^f to an improper fraction, fi — 's?' 
= If, and 109 - 97 = 12; hence, the difference is 12ff. 

In a similar manner, if it is desired to subtract a fraction 
from an integer, reduce 1 to a fraction having the denominator 
of the given fraction and subtract 1 from the integer. Thus, 15 

106. Rule. — Reduce the fractions to fractions having a common 
denominator, find the difference of the numerators, and write the 
remainder over the common denominator; then reduce the resulting 
fraction to its lowest terms. 



EXAMPLES 

(1) Find the difference between || and j%\. Ans. tV/^. 

(2) From 11 A subtract 7§. Ans. 3|i. 

(3) Subtract 8| from 13. Ans. 4|. 

(4) From 4:72^^% take 297xV^. Ans. 175^^. 



58 ARITHMETIC §1 

(5) What number added to 53f will make 751? Ans. 21i|. 

(6) A package of dye weighs 3ji pounds; after If pounds have been 
used how much remains? Ans. Ijf pounds. 



MULTIPLICATION OF FRACTIONS 

107. To Find the Product of a Fraction and an Integer. 

In Art. 53, multiplication was shown to be a short process of 
addition; hence, 4 X f may be regarded as f + f + f + f 

3+3+3+3 4X3 12 - „ .^ . ,^ , -, ,, 

= y = — ~ — = -;=- = If. Here it is seen that it the 

numerator of the fraction be multiplied by the integer, the pro- 
duct is the same as the product of the integer and the fraction. 
This same result may be arrived at in another way, thus : f ths of 
4 is evidently 3 times yth of 4; yth of 4 is 4 -^ 7 = f, and 3 
times 4 one-sevenths is 3 X 4 one-sevenths, or 12 one-sevenths 
= -T- = if ; as before. 

Again, what is the product of 5 and tt? Here, 5 X tt 
~ IT ~ V"= 3f- This same result may be obtained by can- 
celation; thus, ^ X jk = V" = 3f . As the result of this opera- 

3 
tion, it is seen that dividing the denominator of the fraction by 
the integer, multiplies the fraction. This should be evident, 
since the smaller the denominator the larger the fraction; and if 
the denominator is 5 times as small, the fraction must be 5 times 
as large. Therefore, the product of a fraction and an integer may 
be obtained by multiplying the numerator of the fraction by the 
integer or by dividing the denominator of the fraction by the 
integer. If the denominator is not a multiple of the integer, but 
contains one or more factors common to the integer, these factors 
maybe canceled before multiplying the numerator. For example, 
7 

n 

4 

108. To Multiply a Fraction by a Fraction. — When the word 
"of" occurs between two fractions or between a fraction and an 
integer, it has the same meaning as the sign of multiplication; 
thus, I of f and f of 12 have the same meaning as f X f and 



§1 FRACTIONS 59 

I X 12, respectively. Now f is 2 times ^; | of | is 2V, since 1 X i 

is I, and ^ of | must be 3 times smaller, ^ ^ = ^V- Since f is 

5 times i, | of f must be 5 times as large as ^ of |; hence, | of 
8 = T 4 X 5 = A- Since, also, | is 2 times |, and | of | is 3^, 

f of f is A X 2 = i| = A, or p X ^ = t\- 

12 

This same result may be obtained in an easier manner by 

multiplying the numerators of the fractions for the numerator 

of the product, and multiplying the denominators of the fractions 

2X5 
for the denominator of the product; thus, f X f = » .. ^ 

0X0 

_ 1 _ 5 „„ ? V - - ^ - _5 

4 
The product of any number of fractions may be found in this 
same way; that is, divide the product of the numerators by the 

^ H 7 T2 
product of the denominators. For instance, ;s X o X 7^ X 77^ 

' p p jip 17 

4 ^ 
= VV- Cancelation should be employed whenever possible. 

109. To multiply a mixed number by an integer, multiply the 
integral and fractional parts separately. For example, 17| X 42 

7 '^^ 147 
= 17 X 42 + I X 42; g X ^^ = ^ = 36f ; 17 X 42 = 714; 

4 
and 714 + 36| = 750|. 

110. To multiply a mixed number by a mixed number, reduce 
both to improper fractions, and then multiply in the ordinary 

47 25 

manner. Thus, 13f X lOH = ^ X ^ = ^ = 146f 



111. Rule. — I. The product of two or more fractions is found by 
dividing the product of the numerators hy the product of the denomina- 
tors, canceling factors that are common to the numerators and denomi- 
nators whenever possible before multiplying. 

II. To find the product of two or more mixed numbers, reduce 
them to improper fractions before multiplying. 



60 ARITHMETIC §1 



EXAMPLES 

(1) § of f of ' = ? Ans. 



'2 8- 



(2) II X If = ? Ans. A. 

(3) 68 X 19-1 = ? Ans. 1348i 

(4) 33^ X 56| = ? Ans. 1894M. 

(5) 3| X 6f X T^E = ? Ans. llf. 
. (6) m X ^11 = ? ■ Ans. x¥A. 

(7) From a barrel of rosin weighing 404 pounds, xith of the contents 
were taken; how many pounds were taken? Ans. 277| pounds. 

(8) A car load of coal weighing 78,500 pounds was received. At the end 
of a certain period, fths had been used; during a second period, one-half 
of the remainder was used; and during a third period, fths of what was left 
at the end of the second period was burned. How many pounds were used 
during the three periods? Ans. 75,433i| pounds. 

(9) Referring to the last example, how many pounds were consumed 
during each of the three periods? ( First period, 29437^ pounds. 

Ans. \ Second period, 24531| pounds. 
I Third period, 21464|1 pounds. 

(10) What is the value of 4| ounces of dye that is worth $1 j^ per ounce? 

Ans. $5H- 

(11) From a pile of 11,670 cords of wood, Mtbs was used in one part of a 
plant; another part of the plant used }^ as much as the first part, and a third 
part of the plant used K2ths as much as the second part; how many cords 
were used in the third part? Ans. 1276i%2 cords. 

(12) Referring to the last example, how many cords were used in the 

first and second parts of the plant? . f First part, 4376M cords. 

\ Second part, 21883^ cords. 

Note.— Observe that the product of two proper fractions is always less than either 
of the fractions used in finding the product. 



DIVISION OF FRACTIONS 

112. To Divide a Fraction by an Integer. — Suppose it is desired 
to divide | by 3. Since f is 6 one-sevenths, 6 one-sevenths 
divided by 3 is 2 one-sevenths = f , in the same way that 6 
bushels divided by 3 is 2 bushels. Hence, dividing the numerator 
by the integer divides the fraction. Again, the quotient of f 
divided by 3 is evidently only one-third as much as the quotient 
of f divided by 1, since 3 is 3 times 1; hence, the quotient of f 
divided by 3 will be ^ of f = i X f = A = f • Note that in 
the first case, the numerator is divided by the integer, while in 
the second case, the denominator is multiplied by the integer, 



§1 FRACTIONS 61 

the result being the same in both cases. Therefore, dividing 
the numerator or multiplying the denominator by an integer 
divides the fraction. 

113. To Divide an Integer by a Fraction.— Suppose it is desired 
to divide 12 by f; for instance, how many boxes holding f 
pound each can be filled from a package holding 12 pounds? If 
12 be divided by 1, the quotient is 12, but if 12 be divided by a 
number smaller than 1, the quotient will evidently be greater 
than 12. Since 1 is 4 times as large as i, 12 divided by f must 
be 4 times as large as 12 divided by 1; hence, 12 ^ ^ = 12 X 4 
= 48. Since f is 3 times i, 12 divided by f will be only one- 
third as much as 12 divided by i and the quotient will be 48 
X i = 16, or 48 4- 3 = 16 = 12 -- f . 

This same result may be obtained in a much easier manner by 
turning the fraction upside down (this is called inverting the 

4 4 
fraction) and multiplying; thus ;^ X h = 16. Consequently, 

to divide an integer by a fraction, invert the fraction and multiply 
the integer by the inverted fraction. 

114. To Divide a Fraction by a Fraction.— Suppose it is desired 
to divide f by |. Since the divisor is a fraction, invert it and 
multiply I by the inverted fraction; thus, f X f = | = If. 
That this is correct is easily shown. Since f is 3 times i, 3 4- | 
= 3 X f = f, and one-fourth of f is f -^ 4 = |, or f X i = | 
~ Is- 

Example. — Divide j\% by ^. 

• 17 ^ 

SoLUTiON.^Inverting the divisor and multiplying ^ v — = i^ 

= 1t\. Ans. 

115. To divide a mixed number by an integer, divide the inte- 
gral part by the integer; if there is a remainder, annex the fraction 
to it, reduce this new mixed number to an improper fraction, and 
divide it by the integer. 

Example.— Divide 508| by 8. 

Solution.— Here 508 -^ 8 = 63 and 4 remainder: annexing the f to 4 
it becomes 4f = V ; ¥ - 8 = U- Therefore, 508| ^ 8 = 63f i Ans. 

The mixed number might have been reduced to an improper 
fraction and then divided by the integer; the process here given is 
better, however (particularly, when the divisor is small), and it 
requires less work. 



62 ARITHMETIC §1 

116. To divide a mixed number by a mixed number, reduce 
both to improper fractions, and then proceed in the regular 
manner. 

Example. — Divide 15f by 3^%. 

41 5 

Solution.— 15f = i^^; S/tt = f^; ^ X p = W = 4f|. Ans. 

2 21 

117. Rule. — I. To divide a fraction hy an integer, divide the 
numerator or multiply the denominator by the integer. 

II. To divide an integer or fraction by a fraction, invert the 
divisor and multiply the integer or fraction by it. 

m. To divide a mixed number by an integer, divide the integral 
part by the divisor; if there is a remainder, annex the fractional part 
to it, reduce the resulting mixed number to an improper fraction, 
and divide it by the divisor; if there is no remainder, divide the frac- 
tional part of the mixed number by the divisor. The sum of the two 
quotients is the entire quotient. 

Example.— Divide 1296f by 16. 

2 11 
SoLXTTiON.- 1296 ^ 16 = 81; I "^ 16 = § X jg = 24! hence, 1296f 

-j- 16 = Slsi- Ans. That this method of dividing a fraction by an integer 
is correct is easily seen. Since 16 = ^-, inverting the divisor makes it 
Yt, and multiplying by rg^ multiplies the denominator of the fraction by 16. 
This method of dividing by an integer is recommended, because it permits 
of cancelation, when possible. 



EXAMPLES 

(1) Divide f^f by 12. Ans. s\. 

(2) Divide H by 4f . Ans. ^. 

(3) Divide IM by U. Ans. 2f . 

(4) Divide 180 by ||. Ans. 328 

(5) Divide ^ by f . Ans. |. 

(6) Divide 19| by 4|. Ans. 4^. 

(7) Divide 4795i by 21. Ans. 228f. 



COMPLEX AND COMPOUND FRACTIONS 
118. A complex fraction is one that has a fraction for its nu- 

8 3 
merator or denominator or both terms are fractions; thus, «' f 



§1 FRACTIONS 63 

3 

c 
and ^ are complex fractions . In order to distinguish the numerator 

5 
from the denominator, the dividing line is made heavier than 
the dividing line of the fraction or fractions. Thus, in the first 
complex fraction, the numerator is f and the denominator is 7, 
the expression meaning also f -^ 7; in the second complex frac- 
tion, the numerator is 9 and the denominator is f, the expression 
also denoting 9 -r- 5 ; in the third complex fraction, the numerator 
is I and the denominator is f , the expression also denoting f -r- |. 

119. To simplify a complex fraction, multiply the digits or 
numbers above and below the heavy line by the denominator of 
the fraction; the product will be the denominator of the simplified 
fraction in the first and third cases above, and the numerator 
in the second case. Or, transpose the deiiominator of the frac- 
tion fi"om above to helow the line or from below to above the line,as 
the case may be, and use it as a multiplier. For example, to 

3 
simplify the first of the above complex fractions, write 8 

7 

3 3 ■ .u J o = — 5 — = 15 ; in the third 

= n ^ o = -5% ; m the second case, o S 

5 
3 

case, o = o y o = I- That these results are correct may be 

5 

proved by performing the operations indicated. Thus, f -r- 7 

= §__ _ 3 q ^ 3 _ Q X ^ — if^- onrl 3 _^ 3. _ 3 V 5 

7X8~^ ' ^ ~ 3 ~-^' ^"^ 8 ~ 5 — 8 X 3- 

3X8 «• 

Either term or both terms of a complex fraction may consist of 
two or more fractions connected by one or more of the four signs 

11_3 

of operation, +, — , X, -^ ; thus, — j — is a complex fraction, 

, .^ . 1 ^ 112 29 

and it IS equal to -j- = ^^- 



64 ARITHMETIC §1 

120. A compound fraction is a fraction of a fraction. Thus, 
f of TT, or f X TT, is a compound fraction. Since division of 
a fraction by a fraction is changed into multipHcation by in- 
verting the divisor, f -^ yV rnay also be called a compound 
fraction; and a compound fraction may be defined as an expres- 
sion containing two or more fractions connected by signs of 
multipUcation or division. Even the product of one or more 
integers and one or more fractions may be called a compound 
fraction; thus, 145 X If , i of f of 147 = | X f X 147, 3| X 288 
X tV, etc. may be called compound fractions. 

121. If either or both of the terms of a complex fraction is a 
compound fraction, the best way to simplify it is to use the method 

160X^X^^X900X225 

of Art. 119. For example, simplify oqnon 

4 714 3 9 
• ,u , A • . ;0^X38X;WX^00X^^^ 
Transposmg the two denommators ;^ x lf>m X m^^ 

^ m n 

100 

38X714X9 „,,,33 , 
= jQQ = 244lf f . Ans. 

First cancel the ciphers that are common to the numerator and 

denominator; this is the same as dividing by 10, 100, etc. The 

cancelation might have been carried farther, since 38 and 714 

are both multiples of 2, and 100 is a multiple of 4; but it is easier 

to divide by 100 than by 25, and the work was left as it stands. 

291 
As another example, find the value of -3 qi • Reducing 



the mixed numbers to improper fractions, 

4 

v 7 

-y^g-- = o8y2T,- AUS. 



H^ 179 X 3 X 7 



X ¥ 43 X 3 



INVOLUTION 

122. The product of several equal factors is called a power of 
the number used as a factor. Powers are named first, second, 
third, fourth, fifth, etc. according to the number of equal factors 
considered or used. Thus, the second power of 7 is 7 X 7 = 49, 
the third power of 9 is 9 X 9 X 9 = 729, the fifth power of 6 is 
6X6X6X6X6 = 7776. 



§1 INVOLUTION 



65 



123. Instead of indicating the product of the factors as above, 
it is customary to shorten the work by writing a small figure above 
and to the right of the number used as a factor, the small figure 
indicating the number of times the factor is to be used and cor- 
responds with the name of the power. Thus, 6^ is read Q fifth or 
6 to the fifth power, and means the fifth power of 6, or 6 X 6 X 6 
X 6 X 6 = 7776; b"^ is read 5 fourth or 5 to the fourth power, and 
represents 5X5X5X5 = 625, etc. 

The second and third powers are usually called the square and 
cube, respectively; thus, 2V is read 21 square and equals 21 X 21 
= 441, 16^ is read 16 cube and equals 16 X 16 X 16 = 4096. 
The first power of any number is the number itself; thus 217^ 
= 217, 48^ = 48, etc. 

The small figure that is written above and to the right of the 
number is called an exponent. The power itself is the product 
found by performing the multiplication; thus, the cube of 16 is 
4096, the square of 21 is 441, the fifth power of 6 is 7776, the first 
power of 528 is 528, etc. 

Involution treats of powers of numbers. 

124. To indicate the power of a fraction, enclose the fraction 
in parenthesis and write the exponent outside the parenthesis; 
thus, the cube of f is indicated by (f)^, and it equals f X | 

3 _ 3 X 3 X 3 2 7 . p . 

^ ~ 4:X4:X4: ~ ^^' ^^^^^' ^^ ^01^0, a fraction to any 
indicated power, raise the numerator and denominator separately 
to the power indicated. For example, (t^)^ = 

9X9X9X9 7461 . 94 

16 X 16 X 16 X 16 = 65536 '"^ °*^^^ ^°^^^- ^tV)^ = -^.^ 
because A X A X tV X A = -A>19X1^X9__ _ 9^ 
'^ "" 17 16 X 16 X 16 X 16 ~ 16* 

125. To raise 10 to any power indicated by the exponent, 
simply annex to 1 as many ciphers as there are units indicated by 
the exponent; thus, 10^ = 100,000, 10^= 1000, 10^ = 10,000,000, 
etc. In the first case, the exponent 5 indicates 5 units and 5 
ciphers follow 1; in the second case, 3 ciphers follow 1 because the 
exponent is 3; in the third case, 7 ciphers follow 1 because the 
exponent is 7, etc. These results may all be proved by actual 
multiplication. 

Conversely, to express any power of 10 as 10 with an exponent, 
count the number of ciphers, and the number so found will be 
the exponent. Thus, 100 = 10^; 10000 = lO''; 1000000 = 10«; etc. 



66 ARITHMETIC , §1 

To raise a mixed number to the power indicated by the ex- 
ponent, first reduce the mixed number to an improper fraction, 
and then raise the fraction to the indicated power. Thus, 

(4i^)^ = (Ur = -\W- = 22tVt. Ans. 



EXAMPLES 




(2) 7462 = ? 


Ans. 556,516. 


(2) 873 = ? 


Ans. 658,503. 


(3) (fi)3 = ? 


Ans. sSWj. 


(4) 100,000,000 is what power of 10? 


Ans. The 8th. 


(.5) 10^ = ? 


Ans. 10000. 


(6) (291)2 = ? 


Ans. 877|i. 



126. To Multiply or Divide by a Power of 10. — Consider 
what occurs when some integer, as 7035, is multiphed by some 
power of 10, say 10,000. According to Art. 58, the product 
is found by multiplying by 1 and then annexing 4 ciphers, 
the number of ciphers to the right of 1; thus, 7035 
This operation is equivalent to the following: to IQOQO 

multiply any number (integer) by a power of 10, 70350000, 
simply annex to the number as many ciphers as there are ciphers 
in the given power of 10 or as many ciphers as are indicated by 
the exponent of 10. For instance, 10,000 = 10^; hence, to 
multiply any number by 10,000 or 10^, annex 4 ciphers to the 
integer. 

Now any integer, as 7035, may be written 7035., the decimal 
point being always understood to follow the unit figure (5, in this 
case) whether written or not; but when it is written, any number 
of ciphers may be annexed to the number without altering its 
value. Thus, 7035.0, 7035.0000, and 7035 all have the same 
value, since the position of the unit figure has not been changed 
and the addition of the ciphers has not added anything to the 
number. If, now, 7035.0000 be multiplied by 10,000, the product 
will be the same as before, or 70350000., the decimal point being 
moved 4 places to the right. In the product, 5 no longer indicates 
5 units, but 5 ten-thousands. From this, it is evident that any 
number may be multiplied by a power of 10 by moving (shifting) 
the decimal point as many places to the right as there are ciphers 
in the power or indicated by the exponent of 10. Thus, 3.1416 
X 102 = 3.1416 X 100 = 314.16; 3.1416 X 10* = 3.1416 X 



§1 DECIMALS AND DECIMAL FRACTIONS 67 

10000 = 31416; 3.1416 X 10^ = 3.1416 X 10000000 
= 31416000; etc. 

Rule. — To multiply any number by a power of 10, shift the 
decimal point as many places to the right as there are ciphers in the 
power or indicated by the exponent of 10, annexing ciphers, if 
necessary. 

127. Since 3.1416 X 1000 = 3141.6, 3141.6 -^ 1000 must evi- 
dently equal 3.1416; in other words, to divide by a power of 10, 
shift the decimal point as many places to the left as there are 
ciphers in the power or indicated by the exponent of 10. This 
evidently follows also from the fact that division is the reverse of 
multiplication; hence, if the decimal point is shifted to the right 
for multiplication, it must be shifted to the left for division. 

To divide 3.1416 by 10,000, it is necessary to prefix ciphers to 
the first figure, 3, in order to make it occupy a position 4 places 
to the right of unit's place, and 3.1416 -^ 10^ = 3.1416 -4- 10000 
= .00031416. In multiplication, the unit figure is removed to 
the left; in division, it is removed to the right Also, 7035 -^ 10^ 
= 7035 ^ 10000000 = .0007035. 

Rule. — To divide any number by a power of 10, shift the decimal 
point as many places to the left as there are ciphers in the power or 
indicated by the exponent, prefixing ciphers, if necessary. 

It will be noted that the 3 in 3.1416 indicates 3 units, while in 
.00031416, it indicates 3 ten-thousandths, a number ten thou- 
sand times smaller than 3 units. This is evidently correct, since 
the divisor is 10,000. 



DECIMALS AND DECIMAL FRACTIONS 



DEFINITIONS AND EXPLANATIONS 

128. A decimal fraction is one that has a power of 10 for its 
denominator; yV, tWoT; 3tVVA, etc. are decimal fractions. 
When the numerator is divided by the denominator, the above 
fractions become .7, .5236, 3.1416, and are called decimals. 

129. Any decimal may be converted into a decimal fraction by 
writing the decimal (considered as an integer) for the numerator 
of the fraction and writing for the denominator a power of 10 
containing as many ciphers as there are decimal places in the 



68 ARITHMETIC §1 

decimal. For example, .7854 = Am, .00034 = thUhi^, .052465 
~ yoToooiTj etc. 

130. The value of a decimal is not changed by annexing 
ciphers; thus, .25000 = .25. This is evident, since nothing has 
been added to the number represented by the decimal, and the 
position of the first digit relative to the decimal point has not 
been changed. This may also be shown by converting both 

decimals into decimal fractions; thus, .25000 = -^f^f^an^^ — i^' 

and .25 = tVt- Here it is seen that as many ciphers are added to 
the denominator as are added to the numerator, and these may 
be canceled in the fraction. 

131. If the numerator of a decimal fraction contains the factor 
2 or 5, the decimal fraction may be reduced to lower terms, thus 
becoming an ordinary common fraction. For example, .625 

_ 6 25 _ 12 5 _ Si. _ 5. CiOcy — 3 2 _ 4 . 0'7(\A — 2 7 04 

= ill; etc. 

If the decimal does not contain 2 or 5 as a factor, that is,' if it 
does not end in 5 or is not an even number, the decimal fraction 
cannot be reduced, because since 2 X 5 = 10 and they are both 
primes, 2 and 5 are the only factors in any power of 10. Thus, 
1000 = 103 = 10X 10 X 10 =2X5X2X5X2X5 
= 2X2X2X5X5X5= 23 X 53. 

132. Addition and Subtraction of Decimals. — It has already 
been shown how to add decimals and numbers containing 
decimals; simply place the decimal points under one another, and 
add or subtract as in the case of integers, placing the decimal 
point in the result directly under the decimal point in the num- 
bers added or subtracted. The only case that need be considered 
here is that in which the subtrahend contains more decimal 
places than the minuend. In such a case, annex ciphers to the 
minuend until it contains as many decimal places as the subtra- 
hend, and then subtract. 

Example 1.— From 426.45 take 294.0847. 

Solution. — Here the minuend contains two decimal places and the sub- 
trahend contains four; therefore, annexing two ciphers, 
426 4500 which does not change the value of the minuend, the sub- 

294 0847 traction is performed as in the case of integers. This opera- 

tion is equivalent to reducing the decimal fractions to a 
common denominator, which in this case is 10,000. In 
practice, the ciphers would not be written; they would 
simply be considered to be added. 



§1 DECIMALS AND DECIMAL FRACTIONS 69 

Example 2.— From 93 take 77.5652. 

Solution. — The work is arranged as shown in the margin go 

with the decimal points under each other, and the ciphers «y 5652 

that are supposed to be annexed are not written, but are : , 

understood l^-^^^S Ans. 



MULTIPLICATION OF DECIMALS 

133. The only difference between multiplying decimals and 
multiplying integers is the locating of the decimal point in the 
product. The number of decimal places in the product is equal 
to the sum obtained by adding to the number of decimal places in 
the multiplicand the number of decimal places in the multipHer. 
To understand the reason for this, multiply two decimal fractions. 
Thus, .24 X .637 = j\\ X iVA = tVoVA = -15288. Here it 
will be noted that the number of ciphers in the denominator of 
the product is equal to the sum of the number of ciphers in the 
denominators of the two fractions. But the number of ciphers 
in the denominators is the same as the number of decimal places 

.637 in the factors; therefore, multiply the 

■ 24 numbers in the same manner as though they 

2548 were integers, disregarding the decimal points. 

1274 The product is found to be 15288. There are 

. 15288 Ans. 3 decimal places in the multiplicand and 2 in 

the multiplier; hence, there are 3+2 = 5 decimal places in 

the product. 

Had the numbers been .037 and .24, the product as integers 
would have been 888; pointing off 5 decimal places, the product 
as decimals is .00888. 

134. Mixed numbers are multiplied in exactly the same way as 
integers, and the number of decimal places in the product is 
determined in the same manner as in multiplication of decimals. 

Example 1.— What is the product of 75.305 and 3.1416? 

7 5.305 Solution. — The product of the numbers is found 

3.1416 as though they were integers. Since there are 3 

4 5 18 3 decimal places in the multiplicand and 4 in the 

7 5 3 5 multiplier, 7 decimal places are pointed off in the 

3 12 2 product, which is, therefore, 236.578188. 
7 5 3 5 
2 2 5 9 15 



23 6. 5781880 Ans. 



70 ARITHMETIC §1 

Example 2.— Find the product of 47,082 and .0005073. 

4 7 8 2 Solution. — The numbers are multipHed as in 

. 5 7 3 example 1, as though they were integers. The 

14 12 4 6 multiplicand contains no decimal places and the 

3 2 9 5 7 4 multiplier contains 7; hence, 7 decimal places are 

2 3 5 4 10 pointed off in the product. 

2 3.8 8 4 6 9 6 8 Ans. 



EXAMPLES 

(1) 47.95 X 126.42 = ? Ans. 6061.839. 

(2) .0903 X .7854 = ? Ans. .07092162. 

(3) 730.25 X 36.524 = ? Ans. 26671.651. 

(4) 1285 X .0001016 = ? Ans. .130556. 
(6) .00575 X .00036 = ? Ans. .00000207. 

(6) What will be the cost of 7042 tons of coal at $7.33 per ton? 

Ans. $51,617.86. 

(7) If the price of a certain dye is $1.13 per ounce, how much must be 

paid for 18.625 ounces? Ans. $210.4625, or $210.46. 

Note. — In all laonetary transactions, if the fractional part of a cent is equal to or greater 
than J the number of cents is increased by 1; otherwise, the fraction is rejected. In the 
last example, the fraction of a cent is .25 (found by moving the decimal point so as to 
follow the cent), and it is therefore rejected. 

(8) What amount must be paid for 5050 pounds of paper at 12.375 cents 
per pound? Ans. 62,554 cents', or $625.54. 

(9) How much freight will have to be paid on 439 tons of limestone at 
$4.17 per ton? Ans. $1830.63. 

(10) If a pulp mill uses an average of 9.07 tons of soda ash per month 
and the cost is $45 per ton, how much will be paid for this material in a year? 

Ans. $4897.80 



DIVISION OF DECIMALS 

135. Decimals are divided in exactly the same manner as 
integers, no attention being paid to the decimal point until all 
the figures of the dividend have been used. To understand how 
the decimal point is located in the quotient, consider carefully the 
following examples and their solutions, paying strict attention to 
the explanations. 

Example 1.— Divide 7.092162 by .0903. 

Solution. — The division is performed as shown in the y 092162( 0903 
margin, the quotient being 7854 when the dividend and g 321 tqTka 
divisor are considered as integers. To locate the decimal zrr .' 

point, subtract the number of decimal places in the divisor 7004 

from the number of decimal places in the dividend, and the 

remainder will be the number of decimal places in the quo- 4»7d 

tient. In the present case, the divisor contains 4 decimal '*^^^ 

places and the dividend 6; hence, there are 6—4 = 2 3612 

decimal places in the quotient. 3612 



§1 DECIMALS AND DECIMAL FRACTIONS 71 

To understand the reason for above proceeding, convert the 
decimals into decimal fractions and divide by the rule for division 

of fractions Thus ^^^^^^^ ^ -903_ _ 7092162 100^ 

01 iracuons. ±nus, ^qqqqqq • j^qqqq - ioO0000 ^ 903 

= ~Tja?r = 78.54. Here 4 ciphers in the numerator of the divisor 

cancel 4 ciphers in the denominator of the dividend, leaving 6 
— 4 = 2 ciphers in the denominator of the quotient. Since the 
number of ciphers in the denominators of the fractions is the same 
as the number of decimal places in the corresponding numbers 
expressed as decimals, the number of decimal places in the quo- 
tient is always equal to the number of decimal places in the divi- 
dend minus the number in the divisor. 

Example 2.— Divide 7.092162 by 903. 

Solution. — Considering the dividend as an integer, the quotient is 7854, 
as found in the last example. Since there are no decimal places in the 
divisor and there are 6 in the dividend, there are 6—0 = 6 decimal places 
in the quotient, which is therefore equal to .007854. That this is correct 
may be proved by multiplying the quotient by the divisor. 

Example 3.— Divide 70921.62 by .0903. 

Solution. — Considering both numbers as integers, the quotient is 7854. 
Since the divisor contains a greater number of decimal places than the divi- 
dend, annex ciphers to the dividend (this does not change its value, see Art. 
130) until it contains the same number as the divisor; in this case, 2 ciphers, 
then since both dividend and divisor contain the same number of decimal 
places, in this case 4, the quotient is an integer, because 4—4 = 0, and 
there are no decimal places to be pointed off. Therefore, 70921.62 
H- .0903 = 70921.6200 ^ .0903 = 785400, treating the dividend and divisor 
as integers. 

The same result may also be arrived at as follows : Expressing 
the decimals as decimal fractions and dividing as in example 1, 
7092162 10000 7092162 ^^ „ , 

~~1M~ ^ "903 ^ 9Q3 X 100 = 7854 X 100 = 785400. 

Or, since multiplying both numerator and denominator by the 

same number does not alter the value of the fraction, -^^^-^ 

X T¥¥ = ^ Ull^-^, a fraction whose denominator is the same as 

the denominator of the divisor, and which equals the decimal 

70921.6200, which, in turn, is the dividend with 2 ciphers an- 

j ^ ^ 709216200 ;0000 „ ^ 
nexed. But, — JaMa — X qq^ = 785400, the same result as 

before. 



72 ARITHMETIC §1 

Example 4.— Divide 4575.12 by 7.854. 

Solution. — Since the divisor contains 

one more decimal place than the dividend, 

4575 120('7 R'^4 annex a cipher to the dividend. The 

3Q27 — 6 2 A quotient is found to be 582, and there 

582^19 Am. -g ^ remainder of 4092; hence, the 

648 12 Or, 582.521^i9 Ans. q^^tient is 582A|f|. = 582^^, a mixed 

number composed of an integer and a 

19 800 common fraction. It is generally more 

15 708 convenient to express the quotient in 

4 0920 4It4 = tW such cases as a mixed number composed 

3 9270 of an integer and a decimal; in this case, 

16500 simply annex one or more ciphers to the 

157Q8 remainder and proceed with the di- 

TQon vision as indicated. In other words, 

ycr^ tW = .521yt9"' That this is the case 

6 6 _ 1 ^^ readily shown. Thus, 62 = 62.000, 

66 T8^T - rr? and 62.000 ^ 119 = .521y|9. Note 

that three ciphers in all were annexed 

to the remainders, since there were three 

different remainders used as dividends, and one cipher was annexed to each. 

Example 5. — Find to seven decimal places the quotient of xfff f -s"- 

348.0000000(25575 Solution.— Since there are no decimal 

255 75 0136070 Ans. places in either dividend or divisor, annex as 

g2 250 many ciphers to the dividend following the 

^r, ,j2K decimal point as there are decimal places 

required in the quotient, in this case 7. 

When all the figures in the dividend have 
15 3450 

been used, there will be 7 decimal places in 

180000 the quotient, because 7-0 = 7. The 

179025 quotient to 7 decimal places is .0136070. 

9750 

136. Whenever the divisor contains a factor other than 2 or 6, 
there will always be a remainder, no matter to how many decimal 
places the quotient is carried, and as this is usually the case, it is 
customary to carry the division one place farther than the num- 
ber of decimal places specified. If, then the extra figure is 5 or a 
greater digit, the preceding figure is increased by 1, and a minus 
sign is written after the quotient to show that it is not quite so 
large as written or printed; but, if the extra figure is less than 5, it 
is rejected, and a plus sign is written after the quotient to show 
that it is a Kttle larger than written or printed. In this example, 
it is seen that the figure in the 8th decimal place is 3 ; hence, the 
quotient to 7 decimal places would be written .0136070 +. The 
quotient to 3 decimal places would be written .014 — . In prac- 
tice, these signs may be omitted. 



§1 DECIMALS aWd DECIMAL FRACTIONS 73 

Example. — Find the quotient of .0348 4- .00255 to five decimal places. 
.03480 ( .00255 
255 13.647058 

930 13.64706— Ans. Solution. — The divisor contains one more 

765 decimal place than the dividend ; hence, annex- 

TT^Q ing one cipher to the dividend, the integral part 

iKOQ of the quotient is found to be 13. Ciphers are 

■ — — now annexed to the different remainders as de- 
scribed in example 4, and 5 + 1=6 figures of 

the decimal are found. Since the sixth figure 

1°00 Qf ^j^e decimal is 8, the preceding figure, 5, is 

l^o5 increased by 1, and the quotient to five decimal 

1500 places is 13.64706-. 
1275 
2250 
2040 

137. Rule — -I. If the divisor does not contain more decimal 
places than the dividend, divide as though the numbers were integers; 
subtract the number of decimal places in the divisor from the number 
in the dividend, and point off in the quotient as many decimal places 
as are indicated by the remainder. 

II. If the divisor contains a greater number of decimal places than 
the dividend, annex ciphers to the dividend until it contains the same 
number of decimal places as the divisor; if the dividend {with the 
annexed ciphers) is then larger than the divisor, both being considered 
as integers, the quotient when all the figures of the dividend (including 
the annexed ciphers) have been used will be an integer. But, if the 
dividend is smaller than the divisor, annex ciphers to the dividend 
and proceed with the division; the quotient will be a decimal, and 
may be pointed off as in I. 

III. If there be a remainder, it may be written over the divisor, 
and the resulting fraction reduced to lower terms, if possible. Or, 
ciphers may be annexed to the remainders as found and the division 
continued until the quotient has been determined to any desired 
number of decimal places. 



EXAMPLES 

(1) Divide 268.92 by .007007. Ans. 38378.76+ 

(2) Divide 304.75 by 3.125. Ans. 97.52. 

(3) Divide 100.43 by 1000.26. Ans. .100404-. 

(4) Divide 648.433 by 6509.85. Ans. .099608-. 

(5) Divide .00218 by 966. Ans. .000002257-. 



74 ARITHMETIC §1 

(6) The price of a certain dye is $19.25 per pound; if the bill for a purchase 
of dye amounts to 11398.88, how many pounds were bought? 

Ans. 72.67— pounds. 

(7) The amount paid for 3490 tons of coal was $25,642.12; what was the 
price per ton? Ans. $7.35. 

(8) Bought press felts at $2.16 per pound. The bill came to $546.48; 
what was the weight of the felt? Ans. 253 pounds. 



DECIMALS AND COMMON FRACTIONS 

138. To reduce a common fraction to a decimal, divide the 

numerator by the denominator. If the denominator is a power 

of 2, as 2, 4, 8, 16, 32, 64, etc., an exact decimal equivalent can be 

found for the common fraction; this will also occur when the 

denominator is a power of 5, as 5, 25, 125, 625, 3125, etc. But, if 

the denominator contains any other factor than 2 or 5, the quotient 

will never be exact, no matter to how many decimal places the 

division may be carried. In such cases, find the quotient to one 

more decimal place than is desired; if the extra figure is 5 or a 

greater digit, increase the preceding figure by 1 and annex the 

sign — ; otherwise, reject it and annex the sign +. 

Examples. — Reduce (a) xi > (b) ii, and (c) ^f y to decimal fractions : 
Solution. — The work is shown herewith. In (a), the first figure of 

(b) (c) 

17.0 (24 5.000 ( 657 

Ans. 16.8 ■ 70833 + Ans. 4 599 .0076103 + Ans. 
200 4010 

192 3942 

~80 680 

72 657 

~80 2300 

72 1971 



11 

9 


(a) 

.0 (16 
6 . 6875 


1 


40 


1 


28 




120 




112 
80 




80 



8 329 

the quotient evidently denotes .6, since it was necessary to annex one cipher 
to the dividend, 11, before dividing; and since the divisor (denominator) is 
a power of 2 (2* = 16), the quotient is exact, the remaining figures being 
found by annexing ciphers to the different remainders. 

In (b), the first figure of the quotient denotes .7; the remaining figures are 
found as in {a). Note tha,t the second remainder, and all succeeding 
remainders, is 8, and that the third figure of the quotient, and all succeeding 
figures, is 3. 

In (c), it is necessary to annex 3 ciphers to the dividend (numerator) 
before it will contain the divisor (denominator); hence, the first figure 



§1 DECIMALS AND DECIMAL FRACTIONS 75 

of the quotient denotes .007. The division is here carried to 7 decimal 
places. The next figure will be 5, and the correct value of the quotient to 
7 decimal places is .0076104 — . 

139. Mixed Fractions. — It sometimes happens that it is 
desirable to express a quotient exactly, even though it has been 
reduced to an approximate decimal. For instance, in (6) of the 
last example, the quotient might have been written .708/i^ = 
.708^, and the quotient in (c) might have been written .00761 Ay- 
Expressions of this kind are called mixed fractions. 

A mixed fraction may be reduced to a common fraction in the 
same way that a mixed number is reduced to an improper 
fraction; that is, multiply the decimal by the denominator of the 
fraction, add the numerator to the product, and write the 

sum over the denominator. Thus, .708| = •'^Q^ X 3 + -001 

2. 124 +.001 2.125 ^, 
= Q = — o — The numerator, 1, of the fraction be- 
longs to the same order as the figure of the decimal that precedes it ; 
in other words, .7081 = jVW + -Em = .708 + '~ To get rid 

of the decimal in the numerator of the reduced fraction, note 

2 125 X 8 
that .125 = I; hence, multiply both terms by 8, and ' ^ 

= XX A.O,-. nn'7a. 2 3_ _ :00761 X 657 + .00023 



= U' Again, .00761 /A 



657 



_ 4.99977 + .00023 _ ^ 

657 ~ ^^^" 

In practice, the work would be performed as 

.00761 shown in the margin. Here the decimal is 

657 multiplied by the denominator, and the num- 

5327 erator is added to the product before pointing 

3805 off, both the product and the numerator being 

4 566 treated as integers. 

4 99977 

23 140. To Reduce a Decimal to a Fraction 

5 00000 Having a Given Denominator. — To reduce a 

decimal to a fraction having a specified de- 
nominator, all that is necessary is to multiply 
the decimal by the specified denominator; the product will be the 
numerator of the fraction. Thus, to reduce .671875 to a fraction 
having 64 for its denominator, .671875 X 64 = 43; hence, the 
fraction is If. 



76 ARITHMETIC §1 

The reason for this is simple. No number is changed by- 
multiplying it by 1; 1 = If; and .671875 X fi =- wi 

= If. Or, reducing the decimal to a decimal fraction, .671875 
= iVoVVro; riow multiplying both terms of this fraction by 64, 

.^ , , . ^ 671875 X 64 43000000 ^^ 

the specified denommator, j^^^^ ^ ^^ = q^qq^qqq = ^l 

In the case just given, the product of the decimal and the 
specified denominator is an integer, 43. This seldom occurs in 
practice, and what is usually sought is that fraction having the 
specified denominator that is nearest in value to the decimal. 
For instance, what fraction having a denominator of 32 is nearest 
in value to .708? Since .708 X 32 = 22.656, = 23-, .708 
= If, approximately. Had the decimal been .703, .703 X 32 
= 22.496 = 22 +, and .703 = 11 = ii, approximately. Here, 
as before, if the first figure of the decimal part of the product is 5 
or a greater digit, increase the preceding figure by 1 ; otherwise, 
reject the decimal. Had it been required to express .703 as a 
fraction having a denominator of 64, .703 X 64 = 44.992 
., = 45 — , and .703 = ||, approximately. 



EXAMPLES 

(1) Reduce f?f to a decimal. Ans. .80859375. 

(2) Reduce .8035|4 to a common fraction. Ans. fHfH. 

(3) Reduce ^ffx to a decimal. Ans. .0061597 — . 

(4) Reduce xflfo to a mixed decimal. Ans. .0569836^^-. 

(5) Express .8086 inch to the nearest 64th of an inch. 

Ans. It = rl inch. 

(6) Express as decimals |, |, |. Ans. .375, .625, .875. 

(7) Express .3937 to the nearest 12th. Ans. j^^. 

(8) Express .3937 inch to the nearest 32d inch. Ans. ^ inch. 

(9) Reduce .05698ff pound to a common fraction. Ans. yHI^ pound. 



SIGNS OF AGGREGATION 

141. When it is desired to indicate that several numbers are 
to be operated upon as though they were a single number, a sign 
of aggregation is used to enclose the numbers. The word aggre- 
gate means to collect, to bring several things together. There are 
four of these signs : the vinculum , a straight line placed over 
the numbers, and the parenthesis ( ), brackets [ ], and brace { }, 



§1 RATIO AND PROPORTION 77 

used to enclose the numbers. An expression like 3 X (45 — 28) 
or 3 X 45 - 28 means that 28 is to be subtracted from 45 and the 
remainder is to be ' multiplied by 3. In arithmetic, onlj the 
vinculum and the parenthesis are generally used. 

142. Order of Signs of Operation. — When several numbers are 
connected by signs of operation denoting addition and subtrac- 
tion, perform the operations indicated in regular order from left 
to right; thus, 25 - 19 + 47 - 32 - 9 + 50 = 6 + 47 - 32 
- 9 + 50 = 53 - 32 - 9 + 50 = 21 - 9 + 50 = 12 + 50 
= 62. If, however, a sign of multiphcation or division occurs, the 
operation indicated by this sign must be performed before adding 
or subtracting; thus, 25 ~ 3 X 5 + 42 ^ 7 + 38 = 25 - 15 
+ 6 + 38 = 10 + 6 + 38 = 16 + 38 = 54. 

If signs of multiphcation and division follow one another with 
no signs of addition or subtraction between them, perform the 
operations of multiphcation and division in order from left to 
right before adding or subtracting; thus, 46 + 54 -^ 6 X 8 - 67 
= 46 + 9 X 8 - 67 = 46 + 72 - 67 = 118 - 67 = 51. 

Now by using signs of aggregation, the order of operations 
indicated by the signs of operation may be changed; thus, if 
desired, the last expression may be written (46 + 54 -^ q) 
X 8 - 67. Here 54 is first divided by 6, the quotient is added to 46, 
the sum is multiphed by 8, and 67 is then subtracted, the result 
being (46 + 9) X 8 - 67 = 55 X 8 - 67 = 440 - 67 = 373. 
When one sign of aggregation includes another, as in this case, 
always consider the inner sign first. 



EXAMPLES 



(1) 159 - (8 + 5 X 16) -=- 11 - 100 = ? Ans 51 

^2) 6 X 13 - 119 ^ 7-50^ 12 = ? Ans. 5QH- 

(3) 20 + (126 - 4 X 270 -^ 9) X 15 - 35 = ? Ans 75 

(4) 96 - 7.3 X 11 ^ 14 + 24 X 8 = ? Ans. 282".2/j" 



RATIO AND PROPORTION 



RATIO 

143. It is frequently desirable to ascertain the relative sizes of 
two numbers. For instance, suppose it is desired to know the 
relative sizes of 28 and 7, that is, how many times 7 is 28 ? Divid- 



78 ARITHMETIC §1 

ing 28 by 7, 28 -^ 7 = 4; hence, 28 is 4 times 7, or, in other words, 
28 is 4 times as large as 7. What has here been done is to compare 
28 and 7, the comparison being done by division. If it were 
desired to compare 7 to 28, that is, to find what part of 28 is 7, 

divide 7 by 28, obtaining 7 -^ 28 = j; in other words, 7 is 3'^th 

of 28 or 7 is 3'^th as large as 28. 

When two numbers are compared in this manner, by dividing 
the first by the second, the comparison is called a ratio. The 
language used in the first of the above cases is the ratio of 28 to 7, 
and in the second case, the ratio of 7 to 28; in either case, the ratio 
of one number to another is the first number divided by the 
second. 

144. A ratio may be indicated as above by using the sign of 
division, but it is usual to use the colon (see Art. 64), thus indicat- 
ing distinctly that a ratio is implied; thus, the ratio of 57 to 36 is 
written 57 : 36. A ratio is also very frequently indicated in the 
form of a fraction; thus, the ratio of 128 to 16 is written 128 : 16 

128 
or -r^. The second, or fractional, form possesses many advan- 

1 A 

tages. The ratio of 16 to 128 is written 16 : 128 or -=-^. 

The two numbers used in forming or expressing a ratio are 
called the terms of the ratio. The number to the left of the 
colon or above the dividing line is called the first term, and the 
other number is called the second term. In the ratio 32 : 12, 

32 

or Yn, 32 IS the first term and 12 is the second term. 

The value of a ratio is the quotient obtained by dividing the 
first term by the second term; thus, the values of the ratios 57 : 

32 
36, 16 : 128, j^, etc. are liV, h 2|, etc. respectively. 

145. Ratios like the above in which the first term is named first 
in speaking or writing are called direct ratios. It is frequently 
convenient to name the second term first, in which case the ratio 
is called an inverse ratio ; thus, the direct ratio of 24 to 4 is 24 : 4, 
and its value is 6; but the inverse ratio of 24 to 4 is 4 : 24, and its 
value is |. The word direct is seldom or never used in connection 
with ratio; but if the ratio is an inverse one, the word inverse or 
inversely is always used. If there is nothing to -show that the 
ratio is inverse, it is always taken to be direct. 



§1 RATIO AND PROPORTION 79 

The best way of writing an inverse ratio is to write it first as if 
it were direct, and then invert, i.e., transpose, the terms. For 
example, to write the inverse ratio of 36 to 90, write first 36 : 90 

or 7^; now invert the terms and obtain 90 : 36 or 7^. The 
90 36 

value of the inverse ratio of 36 to 90 is 2| = 2.5, 

146. Evidently, only like numbers can be used to make up the 
terms of a ratio. For instance, 5 dollars cannot be compared 
with 8 feet; but 5 dollars can be compared with 8 dollars, and 5 
feet can be compared with 8 feet. The speed of one shaft can be 
compared with the speed of another shaft. 

147. If both terms of a ratio be multiplied or both be divided 
by the same number, it will not alter the value of the ratio. 
Thus, the value of the ratio 32 : 12 is 2f ; multiplying both terms 
by 3, the ratio becomes 96 : 36, and its value is 2§, as before; 
dividing both terms by 4, the ratio becomes 8 : 3, and its value 
is 2f , as before. The reason for this is seen when the ratio is 

32 
written in the fractional form, j^. Now regarding this ex- 
pression as a fraction, it has been shown in connection with frac- 
tions that multiplying or dividing the numerator and denominator 
by the same number does not alter the value of the fraction. 



PROPORTION 

148. If there are two ratios, each having the same value, and 
they are written so as to indicate that the ratios are equal, the 
resulting expression is called a proportion. Thus, the value of the 

ratio o~i^ is f , and the value of the ratid ^^ifa is | ; placing 

these two ratios equal to each other, 5-j = ^■,' ^ , an ex- 

' 8 hours $1.76' 

pression that is called a proportion. 

It is to be noted that the value of any ratio is always an abstract 
number (Art. 6) ; it is for this reason that the ratio of two concrete 
numbers can be placed equal to the ratio of two entirely different 
concrete numbers, as in the above case. Hence, when stating a 
proportion, it is not customary to write the name of the quanti- 
ties, and the above proportion would ordinarily be written either 

as c = YY^ or as 5 : 8 = 1.10 : 1.76. Here 5 is called the first 



80 ARITHMETIC §1 

term, 8 is called the second term, 1.10 is called the third term, 
and 1.76 is called the fourth term. When written in the second 
form above, the two outside terms, 5 and 1.76, are called the 
extremes, and the two inside terms, 8 and 1.10, are called the 
means. 

149. Law of Proportion. — In any proportion, the product of the 
extremes is equal to the product of the means; thus, in the propor- 
tion just given, 5 X 1.76 = 8.8, and 8 X 1.10 = 8.8. 

To understand the reason for this law, the value of the two 
ratios in the above proportion is .625, and the proportion may be 
reduced to the expression .625 = .625. Dividing both numbers 
by .625, the expression reduces to 1 = 1. Now writing the pro- 
portion in the form ^ = y^, divide both ratios by %, and the 

5 8 1 10 8 8 8 

result is g X r = t^ X r, or 1 = ^ = 1. But, 8 X 1.10 

is the product of the means and 5 X 1.76 is the product of the 

extremes, and both are equal to 8.8. 

There is still another way of writing a proportion that was 

formerly used by the older writers on mathematical subjects; 

they employed the double colon in place of the sign of equality 

when writing the proportion in the second form; they 

15 405 
would express the proportion y^ = oKt as 15 : 12 ::405 : 324. 

Here, as before, the product of the extremes equals the product 
of the means, since 15 X 324 = 12 X 405 = 4860. This last 
way of writing a proportion is not much used at this time, the 
sign of equality being preferred. 

150. A proportion may be read in two ways. Consider the 
proportion 16 : 10 = 88 : 55; this may be read 16 is to 10 as 88 
is to 55, or it may be read the ratio of 16 to 10 equals the ratio of 
88 to 55. Either way is correct, but the second is to be preferred 
when using the fractional form for expressing the ratios. 

151. A proportion, like a ratio, may be direct or inverse; in a 
direct proportion, both ratios are direct, but in an inverse propor- 
tion, one of the ratios is inverse. If both ratios were inverse, the 
proportion would become direct again. For example, the state- 
ment that the ratio of 16 tolO equals the inverse ratio of 55 to 88 
indicates an inverse proportion. To write it, first state it as 
though it were a direct proportion and then invert one of the 



§1 RATIO AND PROPORTION 81 

ratios; thus, 16 : 10 = 55 : 88. Now inverting the first ratio, 
10 : 16 = 55 : 88, whence, 10 X 88 = 16 X 55; or, inverting the 
second ratio, 16 : 10 = 88 : 55, whence, 16 X 55 = 10 X 88. 
The proportion as first written was not true, since 16 X 88 
= 1408 and 10 X 50 = 550; but by inverting one of the ratios, it 
became true. The test of any proportion is the law of Art. 149. 
If the product of the extremes does not equal the 'product of the means, 
then the expression is not a proportion. 

152. To Find the Value of One Unknown Term in a Proportion. 

The object of any proportion is to find the value of one of the 
terms when the values of the other three are known; this fact will 
be made clearer shortly. 

Let X represent the value of the term that is not known, and 
suppose the proportion is 14 : 8 = 49 : aj. By the law of pro- 
portion, 14 X a; = 8 X 49. Now, evidently, if 14 times a; = 8 

8 X 49 
X 49, 1 times x must equal — zrg — = 28, and the proportion 

14 : 8 = 49 : 28 is true, because 14 X 28 = 8 X 49 = 392. 
Suppose the first term had been unknown; then re : 8 = 49 : 28, 

8 X 49 
and 1 times x = x = — ^o — = 14. Here it is seen that if the 

unknown is one of the extremes, its value can be found by dividing 
the product of the means by the other extreme. 

Suppose the second term had been unknown; then the propor- 
tion would have been 14 : re = 49 : 28, from which 14 X 28 

14 X 28 
= 49 X re, and the value of x is evidently — j^ — = x, or re = 8. 

Lastly, suppose that the third term had been unknown; then 14: 

14 X 28 
8 = re : 28, and the value of re is evidently -a = re, or re 

o 
= 49. Here it is seen that if the unknown is one of the means, its 
value can be found by dividing the product of the extremes by the 
known mean. 

153. Proportion is used for solving a great variety of problems. 
For example, the statement that "the circumferences of any two 
circles are to each other as their diameters" or that ''the circum- 
feience of a circle varies directly as its diameter" implies a propor- 
tion. In the first statement, if the diameter and circumference 
of any one circle are known and the diameter (or circumference) 
of some other circle is given, the circumference (or diameter) of 
that circle may be found by the method explained in Art. 152. 



82 ARITHMETIC §1 

Thus, the circumference of a circle whose diameter is 5 inches is 
15.708 inches, very nearly; hence, to find the circumference of 
a circle whose diameter is 17.6 inches, first form the proportion 

1 7 6 V 1 5 708 
5 : 17.6 = 15.708 : x, and x = ' ^, = 55.29216 inches. 

o 

Note particularly the way in which the above proportion was 
formed. The first ratio consists of the two diameters, which are 
like quantities, and the second ratio consists of the two circum- 
ferences, which are also like quantities. The first and third 
terms in every direct proportion must belong or relate to the same 
thing, and the second and fourth terms must also belong or relate 
to the same thing; in the present case, 5 and 15.708 are the diame- 
ter and circumference of one circle, and 17.6 and 55.29216 are the 
diameter and circumference of the other circle. It will also be 
noted that the second circle is larger than the first circle ; hence, 
its circumference must also be larger. But, if x be written for 
the third term of the proportion, the value that will be found for 
X will be smaller than the third term, thus showing again that x 
must be written as the fourth term. If desired, the proportion 
might have been written 15.708 '.x = b'. 17.6, and the same 
value will be obtained for x. It is customary, however, to have 
the first ratio consist of numbers or quantities that are known. 

Now referring to the second statement above, that the circum- 
ference of a circle varies directly as its diameter, this means that 
if the diameter increases, the circumference also increases, and 
if the diameter decreases, the circumference also decreases, both 
increasing or decreasing in the same ratio. Therefore, if the 
circumference of a circle whose diameter is 5 inches is 15.708 inches, 
and it is desired to find the circumference of the circle when the 
diameter is increased to 17.6 inches, form the proportion 5 : 17.6 
= 15.708 : X. Substituting the value of x in this proportion, 
5 : 17.6 = 15.708 : 55.29216. The diameter of the second circle 
has increased in the ratio 17.6 : 5, whose value is 3.52, and the 
circumference of the second circle has increased in the ratio 
55.29216 : 15.708, whose value is 3.52; hence, the diameter and 
the circumference increased in the same ratio. From the fore- 
going, it is seen that the word vary, as here used, always implies a 
proportion or that a proportion can be formed. 

In practice, instances frequently occur in which an increase in 
one of the terms of the first ratio results in a decrease in the value 
of the corresponding term in the second ratio, and vice versa. 



§1 RATIO AND PROPORTION 83 

Thus, suppose a certain volume of air be confined in a cylinder, 
and that it is made to keep this volume by means of a piston that 
is free to move up and down and on which is placed a weight. 
Now it is evident that if the weight be increased, the piston will 
move downward and the volume of the air will be less; or, if the 
weight be decreased, the piston will move upward and the volume 
of the air will be greater. In other words, the volume decreases 
as the pressure increases, and vice versa. This fact is expressed 
more elegantly by saying that the volume varies inversely as the 
pressure, which, of course, implies an inverse proportion, when a 
proportion is possible. As an example, it is known that when the 
temperature remains the same, the volume of a gas (air, for 
instance) varies inversely as the pressure; if the volume is 12.6 
cubic feet when the pressure is 14.7 pounds per square inch, what 
is the volume when the pressure is 45 pounds per square inch? 
Forming the proportion as though it were direct, 14.7 : 45 = 12.6 
: X. Here it is seen that the value of x obtained from the pro- 
portion will be greater than 12.6, and it should be less, thus 
indicating an inverse proportion. Since the proportion is inverse, 
invert the terms of one of the ratios, say the second, obtaining 

14 7 V 19 fi 
14.7 :45 = a; : 12.6, from which x - ^^ - 4. 116 cubic 

feet. Note that the volume is smaller than the original volume, 
as it ought to be. 

Example 1. — If 5 men can do a certain piece of work in 22 hours, how long 
will it take 9 men to do the same work, if they all work at the same rate? 

Solution. — It is evident that 9 men can do the work in less time than 
5 men; hence, the proportion is inverse. Stating as a direct proportion, 5 : 
9 = 22: a:; then, inverting the second ratio, 5 : 9 = a; : 22, from which 

5 X 22 ^^-. , ^ 

X = g — = 12^ hours. Ans. 

Example 2. — If 5 men earn $48.95 in 22 hours, how much will 9 men earn 
in the same time at the same rate of pay? 

Solution. — It is here plain that 9 men will receive a greater wage than 5 
men; hence, the proportion is direct, and 5:9 = 48.95 : x, from which 

48.95 X 9 
X = — '—£ = $88.11. Ans. Note that the time, 22 hours, has nothing 

to do with the proportion, because it is the same in both cases. 

154. Compound Proportion. — A simple proportion is one in 
which all the terms consist of but one number in each term; but 
when two of the terms or all four of the terms contain more than 
one number, then the proportion is said to be compounded, or it is 
called a compound proportion. Suppose the last example had 



84 ARITHMETIC §1 

been stated thus: if 5 men earn $48.95 in 22 hours, how much 
will 9 men earn in 15 hours? Here 5 men in 22 hours earn $48.95, 
and it is required to find how much 9 men will earn in 15 hours, 
all being paid at the same rate. The proportion in this case 
should be stated as follows: 5 X 22 : 9 X 15 = 48.95 : x, from 

which X = ot^ — = $60,073^. The reason for stating 

the proportion in this manner is that 5 men working 22 hours is 
the same as one man working 5 X 22 = 110 hours, and during 
this 110 hours, $48.95 was earned. Also, 9 men working 15 
hours is the same as one man working 9 X 15 = 135 hours, and 
during this 135 hours, a certain amount was earned that is re- 
presented by X. Instead of multiplying before substituting in 
the proportion, it is better to indicate the multiplication, so as 
to take advantage of any opportunity of cancelation; thus, 

3 4.45 
9^aiXiW_ 

2 

Example. — If 25 men can dig a ditch 600 feet long, 43^ feet wide, and 3}^ 
deep in a certain number of days, working 8 hours a day, how long a ditch 
4 feet wide and 4 feet deep can 36 men dig when working 10 hours a day? 

Solution. — If men and hours be considered in forming the first ratio, 
the first term is compounded of 25 men and 8 hours, representing time; the 
third term is compounded of 600 feet, 4.5 feet, and 3.5 feet, representing 
work done; the second term is compounded of 36 men and 10 hours; and the 
fourth term is compounded of x feet, 4 feet, and 4 feet. Then, 25 X 8 : 36 
X 10 = 600 X 4.5 X 3.5 : a; X 4 X 4, from which 
_ 36X10X600X4.5X3.5 _ 
"^ ~ 25X8X4X4 ~ ^"^"^^ *^^*- ^''^• 



EXAMPLES 

Find the value of x in the following proportions: 

(1) 6 : 8 = a; : 18. Ans. 13.5. 

^2) 20 : x = 8 : 26. Ans. 65. 

(3) x : 7 = 81 : 91. Ans. 6^3. 

(4) 11 : 16 = 517 : x. Ans. 752. 

(5) 5.5 X 18 : 7.2 X 15 = 4.4 X 25 : 10.5 X^Xx. Ans. 7^. 

'(6) 120 X 64 : 540 X 328 = 110 : a:. Ans. 2536%. 

(7) If a certain supply of provisions will last 525 men 129 days, how long 

will it last 603 men?. Ans. 112 + days. 



§1 RATIO AND PROPORTION 85 

(8) Referring to Art. 153, suppose the volume of an air compressor is 
1.95 cubic feet. If the cyhnder be filled with air at a pressure of 14.7 
pounds per square inch, and the air is compressed until the volume is .31 
cubic feet, what is the pressure, the temperature remaining the same? 

Ans. 92.47— pounds per square inch. 

(9) Referring to Art. 153/. what is the circumference of a circle whose 
diameter is 12% inches? Ans. 38.8773 inches. 



ARITHMETIC 

(PART 2) 



EXAMINATION QUESTIONS 

(1) What is the greatest common divisor of 15,862 and 1309? 

Ans. 77. 

(2) Find the G.C.D. of 45, 135, 270 and 405. Ans. 45. 

(3) Find the L.C.M. of 45, 135, 270 and 405. Ans. 810. 

(4) Which fraction is the larger, ^\ or ^V^? Ans. ^'j;. 

(5) What is the value of f + tV + ii + H - W 

Ans. liW 

(6) Reduce the following fractions to their lowest terms: (a) 
1^1; (&) T2%%', (c) H^; (d) what is the difference between the frac- 
tions (a) and (6) ? (e) what is the sum of the fractions (6) and (c) ? 

Ans. (a) f; (6) ff; (c) ^'r, (d) i^r, (e) ffH. 

(7) What is the value of Ji - ^ + H - li + W 

Jxrib. 5 4 0- 

(8) What is the product of the sum and the difference of 4| 
and3i? Ans. 13AVV. 

(9) A carload of coal weighing 47,960 pounds was delivered to 
a paper mill; Aths was used the first week, yoths the second week, 
and T2ths the third week; how many pounds then remained? 

Ans. 8502 pound. 

(10) What must be paid for 8i| ounces of dye if the cost is 
$1.27 per ounce? Ans. $11.19. 

(11) Divide (a) 0.67 by 0.0007042; (h) 18| by 653.109. 

, r (a) 951.43+. 

^'''- [{h) .028135-. 

(12) The distance across the diagonally opposite corners of a 
hexagonal (six-sided) nut is called the outside diameter, and for 
an unfinished nut is found as follows: to 1| times the diameter of 
the bolt add | inch, and multiply the sum by 1.1547. What is 
the outside diameter of a nut for a 1| inch bolt? Give result to 
nearest 64th of an inch. Ans. 2^\ = 2^2 inches. 

87 



88 ARITHMETIC §1 

(13) Simplify the complex fraction, — «nQ/i >!/ 29 L 52^^ - 

DU54 X TTT ^ T2¥ 

Ans. 6/2. 

(14) The area of a circle is .7854 timet the square of the di- 
ameter; what is the area of a circle whose diameter is 5t\? 

Ans. 21.135+. 

(15) What is the valu e of 6{9 - [14 - 4(12 - 7) (11 + 5) 
-^ 48] + 7(45 - 51 - 13)} ? . Ans, 304. 

(16) Find the value of x in the proportions: (a) 15 : 21 = 
48 : a;; (6) 18 : 8 = a; : 54; (c) 12 : x = 128 : 96. 

((a) X = 67.2. 
Ans. \(b) X =^ 121.5. 
[(c) X = 72. 

(17) If 5 men can do a certain piece of work in 8 days, how 
long will it take 10 men to do 3 times as much work? 

Ans. 12 days. 

(18) If the pressure of steam in an engine cylinder varies in- 
versely as the volume, and the pressure is 110 pounds per square 
inch when the volume is 0.546 cubic feet, what is the pressure 
when the volume is 2.017 cubic feet? 

Ans. 29.78— pounds per square inch. 



AEITHMETIC 

(PART 3) 



SQUARE ROOT 

155. The power of a number has been previously defined; it is 
the continued product of as many equal factors as are indicated 
by the exponent; a very important problem is : given the power, to 
find what equal factor was used to produce the power. The 
number so found is called the root of the given number. The 
roots are named in the same manner as the powers. For example, 
if two equal numbers are used to form the power, one of them is 
called the square root of the given number; if three equal factors 
are used to form the power, one of them is called the cube root; if 
four equal factors are used to form the power, one of them is called 
the fourth root; etc. Thus, the square root of 9 is 3, because 
3 X 3 = 9; the cube root of 125 is 5, because 5 X 5 X 5 = 125; etc. 

156. The root of a number is indicated by writing before it 
what is called the radical sign \/, and to specify what root, a 
small figure is written in the opening of the sign; thus, v^, \/, ^, 
etc. indicate the square root, cube root, and fifth root, respec- 
tively. The square root is indicated so frequently that it is 
customary to omit the small figure, which is called the index of the 
root, using only the sign; hence, \/9, VSl, Vl44, etc. indicate 
respectively the square root of 9, the square root of 81, the square 
root of 144, etc. The index cannot be omitted when indicating 
any other root than the square root. It is also customary to use 
the vinculum in connection with the radical sign. Thus, ^^1728^ 
indicates t he cube root of 1728; v^7776 indicates the fifth root 
of 7776; V24 X 54 indicates the square root of the product of 24 
and 54. If it were desired to indicate the product of 54 and the 
square root of 24, it may be done in two ways, either as 54\/24 
or as V24 X 54. It will be observed that in the first form, 
the sign of multiphcation is omitted between 54 and the radical 
sign; this is customary, and when so written, multiplication is 
always understood. 

89 



90 ARITHMETIC §1 

The root of a number is also frequently indicated by using a 
fractional exponent, the denominator of the fraction indicating 
the root. For example, 196' has the same meaning as \/l96, 
243^ = v^243, 343^ = -^^^343, etc. If the numerator of the frac- 
tional exponent is some number other than 1, it indicates that the 
number is to be raised to the power indicated by the numerator 
and the root taken that is indicated by the denominator. Thus, 
243^ indicates that 243 is to be raised to the fourth power and the 
fifth root is then to be found, or it means that the fifth root of 243 
is to be found and the result raised to the fourth power; the final 
result will be the same. For instance, ■v^243 = 3, and 3* = 81 ; 
also, 343^ = 3,486,784,401 and -^3,486,784,401 = 81, because 
81^ = 3,486,784,401. In other words, 243^ = ( v^243)'or v^243^ 
and the value of both these expressions is 81. Both expressions 
may also be written (243*) ^ or (243^)1 

157. A number is said to be a perfect square, a perfect cube, a 
perfect fifth power, etc., when its square root, cube root, fifth root, 
etc., can be expressed exactly; for instance, 196 is a perfect square, 
because •\/l96 = 14; 343 is a perfect cube, because -v^343 = 7; 
3,486,784,401 is a perfect fifth power, because ^3,486,784,401 
= 81. Only comparatively few numbers are perfect powers ; thus, 
between 1 and 100, the only perfect powers are 1, 4, 8, 9, 16, 25, 
27, 32, 36, 49, 64, 81, and 100, a total of 13 only. Numbers 
that are not perfect powers are called imperfect powers; thus 
7, 12, 26, 47, etc. are imperfect powers. It is to be noted that a 
number may be perfect for some particular power, but not for 
others; in fact, this is usually the case. For example, 243 is a 
perfect fifth power, but it is not perfect for any other power; 16 
= 2^ and 4^, and is a perfect fourth power and a perfect square; 
64 = 2^ = 4^ = 8^, and is a perfect sixth power, a perfect cube, 
and a perfect square. Numbers that are perfect for more than 
one power are rare. 

The root of an imperfect power is never exact; the root of such 
a number may be calculated to any number of figures, in the same 
way that the quotient may be carried to any number of decimal 
places in division when the divisor contains some factor other 
than 2 or 5. 

158. The square root of a number is very frequently desired; 
occasionally, some other root is required, but the process of find- 
ing any root other than the square is so laborious, and such roots 



§1 SQUARE ROOT 91 

are required so seldom, that only the method for finding the 
square root of any number is described here. When other roots 
are desired, they are usually found by using tables of logarithms. 
A method for finding cube and fifth roots is described in Elemen- 
tary Applied Mathematics. 

159. There are many methods of finding the square root of a 
number, the best general method being that known as Horner's 
method, which is the one that will be explained here. 

The first step, no matter what method is used, is to point off the 
number whose root is to be found into periods of two figures each, 
beginning at the decimal point and going to the left, and to the 
right also, in the case of mixed numbers and pure decimals. For 
this purpose, it is best to use a tick — a mark something like an 
apostrophe. For example, the number 11,778,624 is pointed off 
as 1177'86'24; the numbers .7854 and .0003491 would be pointed 
off as .78'54 and .00'03'49'10; the number 755.29216 would be 
pointed off as 7'55.29'21'60. When the right hand period of the 
decimal is not complete, that is, does not contain two figures, add 
a cipher. The left-hand period of the integral part may contain 
either one or two figures, according to whether the number of 
figures in the integral part is even or odd. 

Suppose the square root of 11,778,624 is required. First 

point off the number into periods of 
two figures each. The work is 
done in two columns, as follows: 
The first period, 11, of the given 
number is not a perfect square, and 
the largest perfect square less than 
11 is 9, the square root of which is 
3. Write 3 as the first figure of the 
root, in the same manner as when 
finding the quotient in division, 
and also write it at the head of the 
first column. Now multiply the 
3 in the first column by the 3 in the 
root, and write the product under 
the first period ; subtract, obtaining 
a remainder of 2. Add the first 
figure of the root to the number in the first column, and the sum is 
6, to which annex a cipher, making it 60; also annex the second 
period of the given number to the remainder, 2, in the second 



3 
3 

60 


11'77'86'24(3432 
9 

277 


4 


256 


64 


2186 


4 
680 


2049 

13724 


3 


13724 


683 




3 




6860 




2 




6862 





92 ARITHMETIC §1 

column, obtaining as a result 277. Divide the last number in the 
second column, 277, by the last number in the first column, 60, 
and the quotient is 4, which is probably the next figure of the root. 
Write 4 for the second figure of the root, add it to 60, the last 
number in the first column, obtaining 64 ; multiply 64 by 4, the 
second figure of the root, and subtract the product, 256, from the 
last number in the second column, 277, obtaining 21 for the 
remainder. Then add the second figure of the root, 4, to 64, the 
last number in the first column, and the sum is 68, to which annex 
a cipher, making it 680. Also annex the third period, 86, to the 
remainder 21, in the second column, making it 2186. Divide 
2186 by 680, and the quotient, 3, is probably the next, or third, 
figure of the root. Write 3 for the third figure of the root, add it 
to 680 in the first column, obtaining 683, and multiply 683 by the 
third figure of the root; the product, 2049, is subtracted from 2186, 
and the remainder is 137. Add the third figure of the root, 3, to 
683, obtaining 686, to which annex a cipher, making it 6860. 
Annex the fourth (and last) period of the given power to the last 
remainder, making it 13724; divide this by 6860, and the quotient 
2, is the fourth figure of the root. Add the fourth figure of the 
root to 6860, obtaining 6862 ; multiplying this by the fourth figure 
of the root, the product is 13724, which subtracted from the last 
number in the second column gives a remainder of 0. Therefore, 
\/ll,778.624 = 3432. Ans. / 

Note that the first column is formed entirely by adding, and 
that two additions are made for each figure of the root. The 
above explanation of the method will be made much clearer to 
the reader if he will take pencil and paper and set down the vari- 
ous numbers exactly as they occur in the explanation. After a 
few examples have been solved, the method will be clear, and it 
will be found easy to remember; in fact, the work will be found to 
be but little harder than ordinary division. 

160. Referring to the last example, the numbers in the first 
column (the ones to which the ciphers have been annexed) that 
are used as divisors to determine figures of the root are called 
trial divisors; the numbers that they divide are called trial 
dividends, and the numbers that are obtained by multiplying 
the trial divisors by the root figures may be called, for the want 
of a better term, the partial squares or partial powers. When 
the given number is a perfect square, the sum of the partial 
powers, added as they stand, must equal the given power; thus, 



§1 SQUARE ROOT 93 

the sum of 9, 256, 2049, and 13724, the partial powers, added 

9 as they stand, is 11,778,624, the given number, 

256 In this example, the first trial divisor is 60, 

2049 and the first trial dividend is 277; the second 

^^^^ trial divisor is 680, and the second trial divi- 

11778624 ^gjjjj ig 2186; etc. The numbers obtained by 

adding the root figures to the trial divisors are called the complete 
divisors; 64, 683, and 6862 are the complete divisors. 

When finding the second figure of the root, it may happen that 
the product of the complete divisor and the second figure of the 
root is greater than the partial power; in such case, try a number 
one unit smaller for the second figure of the root. If the product 
of the new complete divisor and the new second figure of the root 
is still greater than the trial divisor, reduce the second figure of 
the root one unit more. It will never be necessary to make more 
than two trials, and seldom more than one. 

Example. — Find the square root of 12,054,784. 

3 12'05'47'84 (3472. Ans. Solution.— The work is almost exactly 

3 9 the same as in the preceding example. It 

60 305 is to be noted, however, that the quotient 

4 256 obtained by dividing the first trial dividend 

64 ^[947 by the first trial divisor is 5; but when 5 is 

^ 4809 added to 60 for the complete divisor, and 

fisn ~r^S4 *^^^ ^^ multiplied by 5, the product is 

7 iQ8«j. greater than the trial dividend; hence, 4 

is tried for the second figure of the root. 

687 

7 
6942 

161. If the given number whose root is to be found contains a 
decimal there will be as many integral places in the root as there 
are periods in the integral part of the given number. Thus, the 
square root of 1205.4784 contains two integral places; the square 
root of 43206.231 contains three integral places, because when 
pointed off, it becomes 4'32'06.23'10, and there are three periods 
in the integral part; etc. 

If the given number is a pure decimal, the root will also be a 
pure decimal; and if there are ciphers between the decimal point 
and the first digit, there will be as many ciphers between the 
decimal point and the first digit of the root as there are periods 
containing no digits. Thus, the square root of .000081 is .009, 
because when pointed off, the given number becomes .OO'OO'Sl, 



94 



ARITHMETIC 



§1 



and there are two periods containing no digits; the square root of 
.000196 is .014, because when pointed off, the given number 
becomes .00'01'96, and there is one period containing no digit; 
etc. To prove this statement, square the roots; thus, .009^ 
= .000081, and .014^ = .000196. 

Example 1. — What is the square root of 844.4836? 

Solution. — On dividing the first trial dividend, 444, by the first trial 
divisor, 40, the quotient is 11; but, the second figure of the root cannot be 

greater than 9; hence, try 9 for the second 



8'44.48'36 (29.06. Ans. 



2 

40 


4 

444 


9 

49 


441 
34836 


9 


34836 


5800 




6 




5806 




Example 2. — Find 


1 


3'56. (18. 8( 


1 
20 


1 

256 


8 


224 


28 


3200 


8 


2944 


360 


25600 


8 


22596 


368 


300400 


8 


264089 


3760 


3631100 


6 


264089 


3766 


22596 


6 


2944 


37720 


224 


7 


1 


37727 


355963689 


7 


36311 


377340 356000000 



figure. The second trial divisor is 580, and 



it is larger than the second trial dividend, 
which is 348; consequently, annex another 
cipher to the trial divisor, making it 
5800, and annex another period to the 
trial dividend, making it 34836, which 
contains the trial divisor 6 times. As 
there are two periods in the integral part 
of the given number, there are two figures 
in the integral part of the root. 
Find the square root of 356 to three decimal places. 

Solution. — The first figure of the root is 1, since 
the first period is less than 4. When dividing 
the first trial dividend by the first trial divisor, 
the quotient is 256 -;- 20 = 12 + ; but the 
second figure of the root cannot exceed 9. 
Trying 9, the complete divisor is 29, and 29 
X 9 = 261, which is greater than the partial 
power, 256; hence, try 8 for the second figure 
of the root. The remainder is 256, showing that 
the given number is an imperfect power. The 
work is now continued by annexing cipher 
periods as shown. The sixth figure of the root, 
the figure in the fourth decimal place, is 9; 
hence, the root correct to three decimal places 
is 18.868—. Adding the partial powers, first 
writing them under the last, 264089, the sum is 
355.963689, which is the square of 18.867; if 
the last remainder, 36311, disregarding the 
decimal point, be added to this sum, the total 
is 356, the given number, showing that the work 
is correct. 

162. If the given number is a common fraction, the root may- 
be found by extracting the square root of the numerator and 
denominator separately; thus, V^ = I- But, unless both 
numerator and denominator are perfect squares, it is better to 
reduce the fraction to a decimal, and find as many figures of the 



§1 SQUARE ROOT 95 

root as are desired; usually, 4 or 5 figures of the root of an im- 
perfect power (not counting ciphers between the decimal point 
and the first digit of pure decimals) are suflficient for practical 
purposes. 

Example. — Find the square root of | correct to four decimal places. 

Solution. — Reducing | to a decimal, it 
9 .87'50 (.93541 Ans. becomes .875. According to Art 161, the 

q oj root is a pure decimal and the nrst ngure 

— — . of it is a digit. After finding the first two 

180 650 figures of the root, there are no more periods 

^-~. .rir to the given number, and the work is con- 

183 10100 tinned by annexing cipher periods to the 

^ 9325 different trial divisors, each period con- 

taining two ciphers. It is evident that 
the fifth figure of the root is 1; hence, the 
root correct to 4 decimal places is .9354. 
To prove that no mistake has been made in 
the work, add the partial powers, obtaining 
.87497316 = .9354^; adding the last re- 
mainder the sum is .875; the given number, 
which shows that no mistake has been 
made. Furthermore, when expressed to 
four figures, .87497316 = .8750, which is 
.87500000 also correct. It is well to check the work 

in this manner, as it prevents mistakes. 

163. If the reader has followed each step of the foregoing 
examples with pencil and paper, he should be able to extract the 
square root of any number. 

Rule I. — Beginning at the decimal point, point off the given 
number into periods of two figures each, including the decimal part, 
if any. 

II. Arrange the work in two columns, the second column contain- 
ing the given number. The first figure of the root is the square root 
of largest square that does not exceed the first period. 

III. Write the first figure of the root as the first number in the 
first column, multiply it by the first root figure, and subtract the 
product from the first period, and annex to the remainder the second 
period; this is the first trial dividend. Add the first root figure ^ to 
the number in the first column, and annex a cipher, thus obtaining 
the first trial divisor. 

IV. Divide the trial dividend by the trial divisor, and the first 
figure of the quotient (if less than 10) will probably be the second 
figure of the root; add this figure to the trial divisor, obtaining the 



1860 
5 


77500 

74816 


1865 
5 


2684( 
.81 


18700 
4 


549 
9325 


18704 


74816 


4 
187080 


.87497316 
2684 



96 ARITHMETIC §1 

complete divisor, which multiply by the second root figure, and sub- 
tract the product from the trial dividend, to which annex the next 
period to form the next trial dividend. But if this product is greater 
than the trial dividend, try a figure one unit less than the quotient 
just obtained, adding it to the trial divisor, and multiplying as before. 
If the product is yet greater than the trial divisor, try a root figure one 
unit less than the last. Having found a satisfactory complete 
divisor, add to it the root figure last found, and annex a cipher to 
form a new trial divisor. Divide the second trial dividend by the 
second trial divisor, and the integral part of the quotient will be the 
next figure of the root. 

V. Proceed in this manner until all the periods of the given 
number have been used. If the number is an imperfect square, 
limit the root to five figures (counting the first digit as one), unless 
more figures are especially desired, annexing cipher periods of two 
figures each, if necessary. 

VI. // at any time, the trial divisor is larger than the trial dividend, 
annex a cipher to the trial divisor and annex another period to the 
trial dividend, placing a cipher in the root. Proceed in this manner 
until the root has been found or as many figures have been found as 
are desired. 

VII. // the quotient obtained by dividing the first trial dividend 
by the first trial divisor is greater than 9, try 9 for the second figure 
of the root. The sum of the partial powers added as they stand, 
must a ways be equal to the square of the root as found, if no mistake 
has been made in the work, and this sum plus the last remainder 
must equal the given number. It is a good plan to check the work 
in every instance by ascertaining if this is the case. 



EXAMPLES 

(1) Find the square root of 293,005.69. Ans. 541.3 

(2) Find \/lO to four decimal places. Ans. 3.1623 — 

(3) Find V3.1416 to four decimal places. Ans. 1.7725- 

(4) Find V36|| to four decimal places. Ans. 6.0557 + 

(5) Find V.02475 to five decimal places. Ans. .15732 + 

(6) Find Vf to five decimal places. Ans. .81650 — 

(7) Find \/.0000000217 to five decimal places. Ans. .00014731- 



§1 SQUARE ROOT 97 

APPLICATIONS OF SQUARE ROOT 

164. In problems relating to mensuration, mechanics, strength 
of materials, and engineering generally, it is required very fre- 
quently to find the square root of numbers. A few applica- 
tions will be given here by means of examples. 

Example 1. — If the area of a circle is known or given, the diameter of the 
circle can be found by multiplying the square root of the area by 1.1284. 
What is the diameter of a circle . having an area of 56.28 square inches? 
Give result to four decimal places. 

Solution. — Since the result is required to four decimal places, and there 
are four decimal places in the number 1.1284 and one integral place, making 
five figures in all, find the square root of 56.28 to 5 + 1 =6 figures, obtain- 
ing 7.50200 -. Then, the diameter is 1.1284X7.502=8.4652568, or 
8.4653 — inches to four decimal places. Ans. 

Example 2. — When a heavy body falls freely, it strikes the ground with 
a velocity in feet per second equal to 8.02 times the square root of the height 
of the fall in feet. If a certain body, say a cannon ball, falls from a height 
of 750 feet, what will be its velocity when it strikes the ground ? 

Solution.— The square root of 750 is 27.386+, and 8.02 X 27.386 
= 219.64— feet per second, or about 2.5 miles per minute. Ans. 

Example 3. — The intensity of light varies inversely as the square of 
the distance of the hght from the object illuminated. If the intensity of 
illumination of a certain object 5 feet from the source of light is 32 candle- 
power, at what distance will the intensity of illumination be 18 candlepower? 

Solution. — Expressed as a direct proportion, 32 : 18 = 5^ : x^. It is 
necessary to square the distances 5 and x, because the intensity varies in- 
versely as the square of the distance. Inverting the terms in the second 

ratio, 32 : 18 = a;2 : 5^, from which x^ = t^ = 44.4444+. Now if 

lo 

^2 = 44.4444, X must equal the square root of 44:A4:4:4:, that is x = a/44.4444 
= 6.6667— feet. Ans. This result might also have been obtained as 
, ., 32 X 25 400 , /400 20 

follows: — Ys — "^^^^'\"9~="3" = ^i = 6.6667- feet. The 

above result means this: if an object is illuminated with an intensity of 32 
candlepower when 5 feet from the light, it will be illuminated with an in- 
tensity of only 18 candlepower when 6| feet from the light. 

Example 4. — The strength of a simple beam (one that is merely sup- 
ported at its ends) varies directly as its breadth, directly as the square of its 
depth, and inversely as its length. If a simple beam made of hemlock is 
20 feet long, has a breadth of 3 inches, a depth of 8 inches, and will safely 
carry a load of 960 pounds, uniformly distributed over the beam, what must 
be the depth of a similar beam having a length of 16 feet and a breadth of 2 
inches to carry safely a uniform load of 1250 pounds? 

Solution. — This is evidently a problem in compound proportion. Stated 
as a direct proportion, 3 X 8^ X 20 : 2 X a;^ X 16 = 960 : 1250. It is neces- 
sary to square 8 and x, because the strength varies as the square of the depth 
7 



98 ARITHMETIC §1 

But the strength varies inversely as the length; hence, transposing 20 and 

16, the lengths, 3 X 8^ X 16 : 2 X a:^ X 20 = 960 : 1250, from which 

, 3X64X16X1250 ,._ -p + vp 2 mn ^TTv^ mi, 

a;2 = 2 V 20 V QBO — ^ ^ "^ ' ^ ^ V 100 = lOj hence, 

the depth of the beam must be 10 inches. Ans. 

The problems just given will suffice to show the importance of 
knowing some method of extracting the square root of numbers. 



PERCENTAGE 

165. It is frequently desirable to compare numbers or 
quantities not in relation to each other, but in relation to some 
fixed number, called the base. For example, the sum of 7, 6, 
and 12 is 25. Now to compare 7, 6, and 12 with 25, form the 
ratios 7 : 25, 6 : 25, and 12 : 25. The values of these ratios are 
found by dividing the first terms by the second, thus obtaining 
■2^ = .28, -is = .24, and if = .48. Note that the sum of the 
fractions is -g^ + A + if = 1, and the sum of the decimal 
equivalents of the fractions is .28 + .24 + .48 = 1.00 = 1, also, 
as it must. In the case of the ratios, the base is 25; but in the 
case of the fractions and decimals, the base is 1, a much more 
convenient number, and one that does not change, as will 
continually be the case with ratios. 

In business transactions, and in many other cases that arise in 
practice, it is convenient to have T^T = .01 as the base, thus making 
the values of the ratios 100 times as large as those obtained above. 
With .01 as the base, the values of the above ratios become 
.28 -f- ^i-Q = .28 X 100 = 28, .24 X 100 = 24, .48 X 100 = 48, 
and 28 + 24 + 48 = 100. In the last case, the three numbers 
added are called 28 per cent, 24 per cent, and 48 per cent, 
respectively, of 25, which is 100 per cent of 25. 

The term per cent is an abbreviation of the Latin words per 
centum, which mean by the hundred, and percentage means a cer- 
tain number of hundredths of some number. Thus, 28 per cent 
of 25 is 25 X TW = 25 X .28 = 7, and 7 is the percentage 
obtained by multiplying 25 by 28 hundredths. Also, 7 per cent 
of 438 is 438 X t^ = 438 X .07 = 30.66, and 30.66 is the 
percentage obtained by multiplying 438 by 7 hundredths. 

166. The sign or symbol that is used to denote per cent is % ; 
thus, 7% is read 7 per cent, 113^^% is read 11% per cent, etc. 
The number that is placed before the symbol and shows how 



§1 



PERCENTAGE 



99 



many hundredths are to be taken or considered is called the 
rate per cent; thus, in the last sentence, 7 and ll}i are rates per 
cent. When the rate per cent is expressed decimally or frac- 
tionally, thus denoting its actual value, it is called the rate. For 

instance, if the rate per cent is 5%, the rate is .05 or ^^; if the 

no 7 KQ7 

rate per cent is 58.7%, the rate is .587 or ^ = ^ , etc. 

The rate, therefore, is always equal to the rate per cent divided by 
100; and the rate per cent is 100 times the rate. 

167. The rate can frequently be expressed as a simple fraction 
by reducing the common-fraction form of the rate to its lowest 

terms. For instance, 6i% = ^ = ^ = -,V, or 6|% = ^ 



625 
10000 



— tV- In the table below, some of the rates per cent 

commonly used, with their decimal and common fraction equiva- 
lents, are given. 



Rate Per Cent 


Rate 


Rate Per Cent 


Rate 


11 


.015 =^t^ 


25 


.25 = i 


2 


•02 =,V 


33f 


.331= i 


3i 


■03^ =,V 


371 


.375= f 


4 


.04 =^V 


40 


•4=1 


41 


•04i =A 


50 


.5 = i 


5 


.05 =,V 


62i 


.625= 1 


6i 


.0626 =^V 


661 


.661= 1 


8 


•08 =A 


75 


.75 = 1 


10 


•1 =^^ 


83i 


•831= 1 


12i 


.125 =1 


87i 


.875= 1 


161 


.16! =i 


100 


1 =1 


20 


•2 =i 


250 


2.5 =2* 


1 


•00125 = 3.1^ 


475 


4.75 =4| 



168. The majority, if not all, of the fractions in the above table 
may be used to advantage in computing the percentage instead 
of their decimal equivalents. For example, to find 12>^% of 
35.164, simply divide 35.164 by 8, obtaining 4.3955; this is cor- 
rect, since 121.^% = |, and 35.164 X i = 35.164 ^ 8. Simi- 
larly, for the same reasons, 83>^% of 439.2 = 439.2 = ^ 

439.2 X 5 
= -Q = 366. It is evidently easier to multiply by the 



100 ARITHMETIC §1 

fractions than it is to multiply by their decimal equivalents, and 
there is less liability of making mistakes. 

169. The base may now be defined as the number that is 
multiplied by the rate to obtain the percentage. In the first 
case of Art. 168, 35.164 is the base and 4.3955 is the percentage; 
in the second case, 439.2 is the base and 366 is the percentage. 
Consequently, 

Rule. — To find the percentage, multiply the base by the rate. 

Example. — Out of a lot of 1500 reams of paper, 37% were sold the first 
month and 43% were sold the second month; how many were sold in the two 
months? 

Solution. — This problem may be solved in two ways : 1st method. The 
number of reams sold the first month is 1500 X .37 = 555; number sold the 
second month is 1500 X .43 = 645; number sold in the two months is then 
555+645 = 1200. Ans. 2d method. If 37% were sold the first month and 
43% were sold the second month, 37% + 43% = 80% were sold in the two 
months; and 1500 X .80 = 1200 reams were sold in the two months. Ans. 

170. If the product of the base and the rate equals the percent- 
age, the rate must equal the percentage divided by the base, since 
the product of two factors divided by one of the factors must give 
the other factor for the quotient. Therefore, 

Rule. — To find the rate, divide the percentage by the base. The 
rate per cent is 100 times the rate. 

Example. — Out of a lot of 1500 reams of paper, 555 were sold; what per 
cent of the paper was sold ? 

Solution. — Here 555 is the percentage and 1500 is the base, since it is the 
number which, multiplied by some rate, gives a product of 555. Hence, 
555 -^ 1500 = .37 = the rate, and .37 X 100 = 37, the rate per cent. 
Therefore, the per cent of paper sold was 37%. Ans. 

171. If the rate and percentage are known and it is desired to 
find the base, divide the percentage by the rate. This is evi- 
dently correct, since, if the product of the base and rate equals 
the percentage, the percentage divided by the rate must equal the 
base. Consequently, 

Rule. — To find the base, divide the percentage by the rate. 

Example. — In a certain month, 43% of a certain lot of pulp was sold; 
if the number of bales sold was 645, how many bales were in the lot? 

Solution. — Here the rate is .43 and the percentage is 645; hence, the base 
is 645 -f- .43 = 1500, the number of bales in the lot. Ans. 

172. It is to be noted that the rate per cent and the percentage 
really represent the same thing. The rate per cent is the number 
of parts in 100 parts of the base, while the percentage is the actual 



§1 PERCENTAGE 101 

number of parts of the base. In other words, the rate per cent 
is found on the assumption that the base is always 100, while 
the percentage is computed on the actual value of the base. 

173. There are two other terms used in percentage — the amount 
and the difference. The amount is equal to the sum of the base 
and percentage, the difference is equal to the remainder obtained 
by subtracting the percentage from the base. To illustrate the 
meaning of these terms, suppose that the price of alum is $42.00 
per ton, and that the price was increased 123-^%; what is the new 
price? $42 -T- 8 = $5.25 = 12>^% of $42.00; then the new price 
is evidently $42.00 + $5.25 = $47.25 = the amount, since it is 
equal to the sum of the base ($42.00) and the percentage ($5.25). 
Again, suppose, as before, that the price is $42.00, but that the 
price is reduced 123^-2%, what is the new price? Since 123-^% of 
$42.00 is $5.25, the new price is $42.00 - $5.25 = $36.75 = the 
difference, since it is equal to the base minus the percentage. 

174. The amount may be found in an easier way. Since the 
base alwaj^s represents 1 or 100%, the amount may be represented 
by 1 + rate or by 100% + rate per cent; hence, if the base and 
rate are given, the amount may be found by multiplying the 
base by 1 + rate. Thus, referring to Art. 173, the new price 
after the increase is $42 X (1 + .125) = $42 X 1.125 = $47.25, 
since 123^^% = .125. Therefore, 

Rule. — The amount equals the base multiplied by 1 plus the rate. 

Example. — If a man's wages is $3.20 per day, and he receives an increase 
of 15%, how much does he then receive per day? 

Solution. — The base is $3.20, the rate is .15, and the new rate of wages 
is the amount, which is $3.20 X 1.15 = $3.68. Ans. 

175. If the amount and the rate are known, and it is desired to 
find the base on which the rate was computed, divide the amount 
by 1 plus the rate. This is evidently correct, since the amount is 
equal to the base multiplied by 1 plus the rate. Hence, 

Rule. — To find the base, divide the amount by 1 plus the rate. 

Example. — If a man receives $3.68 cents per day in wages, and this is 
15% more than he formerly received, how much did he formerly receive? 

Solution. — Here $3.68 is some number plus 15% of that number; that 
is, $3.68 is the amount and 15% is the rate per cent. Therefore, he formerly 
received $3.68 -v- 1.15 = $3.20. Ans. 

176. If the base and amount are given and it is desired to know 
the rate, divide the amount by the base and subtract 1 from the 
quotient. This is evidently correct, since dividing the amount 



102 ARITHMETIC §1 

by the base gives a quotient that equals 1 plus the rate, and sub- 
tracting 1 from this leaves the rate. 

Rule.— To find the rate when the base and amount are known, 
divide the amount by the base and subtract 1 from the quotient. 

Example. — If a man's wages are $3.68 per day and he formerly received 
$3.20 per day, what rate per cent, increase did he receive? 

Solution. — Here $3.)68 is the amount, since it equals $3.20 plus a certain 
percentage of $3.20, the base. Consequently, 3.68 -f- 3.20 = 1.15; 1.15 — 
1 = .15, the rate; and .15 X 100 = 15%. Ans. The solution may also be 
obtained as follows : $3.68 — $3.20 = $0.48, the actual increase per day that 
he received, and which is the percentage. Hence, the rate is .48 -i- 3.20 
= ,15, and the rate per cent is .15 X 100 = 15%. Therefore, also. 

Rule. — To find the rate when the base and the amount are known, 
subtract the base from the amount and divide the remainder by the base. 

The second rule is rather easier to remember and apply than 
the first rule. 

177. The rules just given concerning the amount may also be 
used when the difference is given, by subtracting the rate instead 
of adding it. Thus, referring to Art. 173, the new price after the 
old price had been reduced 12>^% is $42 X (1 - .125) = $42 X 
.875 = $36.75. Therefore, 

Rule. — To find the difference, multiply the base by 1 minus the 
rate. 

In connection with problems relating to money and prices, the 
amount deducted from a fixed price or amount (the base) is 
called the discount. Insofar as percentage is concerned, the 
words percentage and discount mean the same thing, when finding 
the difference. 

Example. — If a certain article is priced at $4.75 and it is sold at a discount 
of 8%, how much was received for it? 

Solution. — The base is $4.75, the rate is .08, and the difference is re- 
quired. Applying the rule, $4.75 X (1 - .08) = $4.75 X .92 = $4.37. 
Ans. The solution may also be obtained as follows: $4.75 X .08 = $0.38, 
the discount; $4.75 — $0.38 = $4.37. Ans. Either method may be used, 
whichever appears to be easier. 

178. If the difference and rate are known, and it is desired to 
find the base, divide the difference by 1 minus the rate. For, 
since the product of the base and 1 minus the rate equals the 
difference, the base must equal the difference divided by 1 minus 
the rate. Consequently, 

Rule. — To find the base when the rate and difference are known, 
divide the difference by 1 minus the rate. 



§1 PERCENTAGE 103 

Example. — $4.37 was paid for a certain article that was bought at a dis- 
count of 8%; what was the original price of the article? 

Solution. — Here $4.37 is the difference and .08 is the rate; the base = 
the original price = $4.37 -^ (1 - .08) = $4.37 -i- .92 = $4.75. Ans. 

179. If the difference and base are known, and it is desired to find 
the rate, subtract the difference from the base and the remainder 
will be the discount; divide the discount by the base, and the 
quotient will be the rate. 

Example. — If an article is bought for $4.37 and the original price was 
$4.75, what was the per cent of discount received? 

Solution.— The actual discount received is $4.75 - $4.37 = $0.38, and 
$4.75 is the base; hence, 0.38 -^ 4.75 = .08 = 8% of the original price = 
per cent of discount. Ans. 

180. Chain Discount. — Certain manufacturers that deal in 
goods that are subject to rapid changes in prices issue cataloges 
in which the prices are quoted at a much higher figure than they 
think will ever be asked. They then issue what are called dis- 
count sheets, which are quickly printed, and are subject to change, 
in accordance with the state of the market. When more than 
one discount is quoted on any one article, it is called a chain 
discount. In computing a chain discount, that quoted first is 
deducted from the list price, the second discount is deducted from 
the remainder, the third discount from the last remainder, etc. 
It is never allowable to add the rates of discount and then deduct. 

As an example, suppose that the price (catalog) of a certain 
article is $26.60, and that discounts of 60,10 and 5% are offered; 
what is the real price of the article? 

$26.60 X (1 - .60) = $10.64, the price after 60% has been 
deducted. $10.64 X (1 - .10) = $9,576, or $9.58, the price 
after deducting 60 and 10%. $9.58 X (1 - .05) = $9,101 or 
$9.10, the price after all the discounts have been taken out. 

This same result may be obtained by multiplying $26.60 by 
the continued product of 1 minus the different discounts, that is 
by (1 - .60) X (1 - .10) X (1 - .05) = .40 X .90 X .95 = 
.342; thus, $26.60 X .342 = $9.0972, or $9.10, as before. The 
real discount in this case is, therefore, 100 — 34.2 = 65.8%, 
which may be called the equivalent discount, and 1 — .342 = 
.658 may be called the equivalent discount rate. 

Rule. — To find the equivalent discount rate for any chain dis- 
count, subtract all the discount rates from 1 and find their product; 
then subtract this product from 1. The equivalent discount equals 
the equivalent discount rate multiplied by 100. The net price or 



104 ARITHMETIC §1 

net cost may be obtained by subtracting all the discount rates from 1, 
finding their product, which multiply by the given price or cost. 

Example. — A bill of goods amounting to $102.04 is subject to a chain 
discount of 87 J^, 10, 5, and 2}^%; what is the equivalent discount per cent 
and how much must be paid to settle the bill? 

Solution. — The equivalent discount rate is 1 — (1. — .875) X (1 — 10) 
X (1 - .05). X (1 - .025) = 1 - .125 X .9 X .95 X .975 = 1 - .104203125 
= .895796875, or 89.5796875%. Ans. The amount to be paid to settle 
the bill is $102.04 X .104203125 = $10.63+. Ans. 

181. Gain or Loss Per Cent. — If an article is bought for a 
certain amount and is sold for another amount, there is a gain or 
loss equal to the difference between the two amounts. To find 
the gain or loss per cent, divide the gain or loss by the cost. Thus, 
if a manufacturer finds that it costs him $2.43 to turn out a 
certain article, and he sells it for $3.15, he gains $0.72, and the 
gain per cent is .72 ^ 2.43 = .2963; .2963 X 100 = 29.63%. 
But if he had sold it for $2.25, he would have lost $2.43 - $2.25 
= $0.18, and the loss per cent. = .18 -^ 2.43 = .0741;. 0741 
X 100 = 7.41%. 

Again, suppose that the transmission system of a certain power 
plant is found to absorb 4.35 horsepower, but after making certain 
changes, it absorbs only 3.08 horsepower; what is the saving in 
per cent of horsepower wasted in transmission? The saving in 
horsepower is 4.35 — 3.08 = 1.27; the horsepower wasted before 
the change is the base in computing the per cent; hence, 1.27 -^ 
4.35 = .292- ; and .292 X 100 = 29.2%. In other words, 29.2%, 
of the power formerly wasted is saved by the change. 

In calculations of this kind, it is sometimes a little difficult to 
determine which of the two given numbers is the base; this can 
always be determined correctly by remembering that a gain or 
loss involves a change of some kind; in the case of buying and 
selling, the selling is a change (of ownership) ; in the case of the 
power plant, 4.35 horsepower was changed to 3.08 horsepower. 
The number used to divide the gain or loss is the number that is 
changed; in the first illustration above, the number changed was 
$2.43, and in the second illustration it was 4.35. 

Rule. — To find the gain or loss per cent, divide the gain or loss 
by the number that is changed and multiply the quotient by 100. 

Example 1. — Suppose the indicated horsepower of a certain non-condens- 
ing steam engine is 168, and that it is 204 after a condenser has been attached. 
What is the gain per cent in horsepower? 

Solution. — The horsepower before the change was 168, and after the 



§1 PERCENTAGE 105 

change it was 204; the gain is 204 - 168 = 36; 36 -^ 168 = .2143, and 
.2143 X 100 = 21.43%, the gain per cent. Ans. 

Example 2. — A certain business is valued at $234,517.00; three years 
later, it is valued at $187,250; what was the depreciation in per cent? 

Solution.— The actual depreciation was $234,517 - $187,250 = $47,267; 
the value before the change was $234,517; 47,267 -r- 234,517 = .2016-; 
hence, the depreciation in per cent was .2016 X 100 = 20.16%. Ans. 



EXAMPLES 

(1) What is 4%% of $683.32 to the nearest cent? Ans. $32.46. 

(2) If 475 is 623^% of some number, what is the number? Ans. 760. 

(3) What is %% of $4521? Ans. $16.95. 

(4) What per cent of 840 is 119? Ans. 14K %• 

(5) If an automobile was bought for $1350 and after being used for a year 
was sold for $925, what was the loss per cent? Ans. 31.48%. 

(6) Bought some dye stuff for $21.85, receiving a discount of 5% from the 
selling price; what was the selling price? Ans. $23.00. 

(7) A machinist turned 47 bolts in one day, which was almost 7% more 
than he tiirned the day before; how many did he turn the day before? 

Ans. 44. 

(9) Bought a line of supplies, the bill amounting to $428.73 and subject 
to a discount of 333^3, 10, and 3% if settled within ten days; how much is 
required to settle in ten days? Ans. $249.52. 

(10) A test showed that the coal consumption in the boiler used to furnish 
steam for a non-condensing engine was 660 pounds per hour; after attach- 
ing a condenser, only 541 pounds of coal per hour were required. What was 
the saving per cent? Ans. 18 + %. 

(11) The average daily production of a paper mill in 1917 was 94 tons; 
in 1918, it was 105 tons; what was the per cent of increase? Ans. 11.7%. 

(12) A mill was using 1.4 cords of wood per ton of paper; as the result of 
certain changes, a saving of 4% of wood was effected. What amount of 
wood was then used per ton? Ans. 1.344 cords per ton. 

(13) A pulp mill increased its production 43%, producing 160 tons per 
day; how much did it formerly produce? Ans. 112 tons per day. 

(14) If 165 pounds of clay is used in an order of 2240 pounds of paper, 
what per cent of the finished product is clay?^ Ans. 7.37%. 

(15) A sample of paper weighs 9.372 grams, of which 0.621 grams are 
water and 8.751 grams are fiber; what is the percentage of water and fiber 
in the paper expressed as per cent? ylns. /6.63% of water. 

\ 93.37% of fiber. 

(16) A sample of coal shows 11.3% ash; assuming that all the combustible 
part of the coal is consumed, how many pounds of ashes must be handled 
per ton of 2240 pounds of coal? Ans. 253.12 pounds. 

(17) After having been in use for some time, it was found that a 58-foot 
belt had stretched 3 % ; what was the length of the belt after being stretched? 

Ans. 59.74 feet. 



106 ARITHMETIC §1 

(18) A paper shrinks 1.62% from wet end to calenders; if it was 164 
inches wide on the wire, how wide is it when dry V Ans. 161.34 inches. 

(19) How much bone dry fiber is contained in 793 tons of wet pulp, the 
moisture content being 43 % ? Ans. 452 tons, 

(20) The bone dry weight of pulp is 90% of the air dry weight, which is 
the basis of payment. Referring to example 19, what is the equivalent air 
dry pulp? Ans. 502.22 tons. 

COMPOUND NUMBERS 

182. A compound number is one that requires more than one 
unit to express it; thus, if the length of a piece of hose is stated to 
be 12 feet 7 inches, 12 feet 7 inches is a compound number, because 
two units — the foot and the inch — are required to express the 
length. Had the length been stated as 12 %2 feet, the number 
would be called a simple number, since only one unit — 'the foot — 
is required to express it. 

Compound numbers are usually denominate numbers, a 
denominate number being one having its unit or units de- 
nominated, or named. 

183. A denominate number is always a concrete number (see 
Art. 7), and when only one unit is named, as 5 yards, 8 pounds, 
etc., it is a simple number. An abstract number is usually a 
simple number, but it may be treated as a compound denominate 
number if desired. Thus, 364 may be regarded as 3 hundreds 6 
tens 4 units. 

184. In the case of abstract numbers, each figure belongs to a 
denomination that is ten times as large as the denomination of 
the next figure on its right and one-tenth as large as the denomi- 
nation of the next figure on its left. In connection with de- 
nominate numbers, however, this is seldom or never the case. 
For instance, in the compound denominate number 4 yards 2 
feet 8 inches, it takes 12 inches to make 1 foot, the next higher 
denomination, and it takes 3 feet to make 1 yard, the next 
denomination higher than feet. For this reason, it is necessary 
to memorize certain tables showing the relation of the different 
denominations for different compound numbers. 

In order to save time and space in writing, the names of the 
units are abbreviated, and these abbreviations are given in the 
tables that follow. 

185. The tables that follow should all be thoroughly committed 
to memory. Only the tables in common use are given here. 



§1 COMPOUND NUMBERS 107 

Under each table is a subsidiary table that shows the relation 
between the different units; it is not necessary to memorize these, 
although it will be found convenient to remember some of the 
principal equivalents, as, for example, that 5280 feet make a 
mile, that 36 inches make a yard, etc. 

TABLE I 

LINEAR MEASURE 

12 inches (in.) = 1 foot ft. 

3 feet = 1 yard yd. 

5)4 yards = 1 rod rd. 

40 rods = 1 furlong fur. 

8 furlongs (320 rd.) =1 mile mi. 

mi. fur. rd. yd. ft. in. 

1 = 8 = 320 = 1760 = 5280 = 63,360 

1 = 40 = 220 = 660 = 7,920 

1 = 5.5 = 16.5 = 198 

1 = 3 = 36 

1 = 12 

186. Another abbreviation for feet and inches, much used by 
draftsmen, mechanics, etc. is (') for feet and (") for inches; hence, 
4 feet 8 inches may be written either 4 ft. 8 in. or 4' 8". When 
the latter form is used, it is advisable to place a dash between the 
feet and inches; thus, 4' — 8". The subsidiary table is useful in 
expressing higher units in terms of the lower units, and vice versa. 
For example, if it were desired to express 37 rd. in inches, the 
table shows that 1 rd. = 198 inches; hence, 37 rd. = 37 X 198 
= 7326 in. Again, what fraction of a mile is 1360 feet? From 

the table, 1 mi. = 5280 ft.; hence, 1360 ft. = ^^^on ^ ^^ ^^• 

= .2576 - mi. 

TABLE II 

SQUARE MEASURE 

144 square inches (sq. in.).. . = 1 square foot sq. ft. 

9 square feet = 1 square yard sq. yd. 

30 J^ square yards . , = 1 square rod sq. rd. 

160 square rods = 1 acre A. 

640 acres = 1 square mile sq. mi. 

sq. mi. A. sq. rd. sq. yd. sq. ft. sq. in. 

1 = 640 = 102,400 = 3,097,600 ^ 27,878,400 = 4,014,489,600 

1 = 160 = 4,840 = 43,560 = 6,272,640 

1 = 30.25= 272.25 = 39,204 

1 = 9 = 1,296 

1 = 144 



108 ARITHMETIC §1 

187. A plot of ground in the form of a square, each side of which 
measures 208.71 feet, say 208 ft. 9 in., contains one acre. 

TABLE III 

CUBIC MEASURE 

1728 cubic inches (cu. in.).. . . = 1 cubic foot cu. ft. 

27 cubic feet = 1 cubic yard cu. yd. 



128 cubic feet = 1 cord (wood) cd. 

24^ cubic feet = 1 perch (stone, masonry)?. 

cu. yd. cu. ft. cu. in. 

1 = 27 = 46,656 

1 = 1,728 

188. The cord is used only in measuring wood. A pile of wood 
8 ft. long, 4 ft. wide, and 4 ft. high contains one cord. The perch 
is used in measuring stone and brick walls and other masonry. 
Some contractors allow 25 cubic feet to the perch, but 24 % 
cubic feet is the correct value. 

TABLE IV 

AVOIRDUPOIS WEIGHT 

4373^ grains (gr.) = 1 ounce oz. 

16 ounces = 1 pound lb. 

100 pounds = 1 hundredweight .... cwt. 

20 hundredweight (2000 lb.) = 1 ton T. 

T. cwt. lb. oz. gr. 

1 = 20 = 2000 = 32,000 = 14,000,000 

1 = 100 = 1,600 = 700,000 

1 = 16 = 7,000 

1 = 437.5 

189. The ton of 2000 pound, called the short ton, is the one 
commonly used. The ton of 2240 pounds is called the long ton, 
and is used to weigh coal, pig iron, and other coarse commodities; 
it is the basis of freight rates on foreign exports. In connection 
with the long ton, 14 pounds make a stone, 2 stones make a 
quarter, 4 quarters make a hundredweight, and 20 hundred- 
weight make a ton; hence. 









LONG 


TON 








T. 


cwt. 




qr. 




St. 




lb. 


1 = 


= 20 


= 


80 


= 


160 


= 


2240 




1 


= 


4 


= 


8 


= 


112 








1 


^ 


2 
1 





28 
14 



§1 



COMPOUND NUMBERS 



109 



190. For measuring medicines, jewelry, gold, silver, etc., the 
Troy ounce, which contains 480 grains is used. The Troy pound 
contains 12 Troy ounces or 5760 grains. The Troy pound is 
therefore, only |^^t = iflth of an avoirdupois pound. A pound 
of gold thus weighs 5760 -^ 437.5 = 13tS^ = 13.1657 + avoir- 
dupois ounces. 



TABLE V. 

LIQUID MEASURE 



4 gills (gi.) 

2 pints 

4 quarts 

313^ gallons 

2 barrels (63 gallons) . 



hhd. 
1 



bbl. 
2 
1 



gal. 
63 
31.5 
1 



= 1 pint pt. 

= 1 quart qt. 

= 1 gallon gal. 

= 1 barrel bbl. 

= 1 hogshead hhd. 

qt. pt. gi. 

252 = 504 = 2016 

126 = 252 = 1008 

4 = 8 = 32 

1 = 2 = 8 

1 = 4 



191. The United States, or wine, gallon contains 231 cubic 
inches, and a gallon of water weighs very nearly 8^ pounds. A 
cubic foot contains 1728 ^ 231 = 7.481 gallons, or, roughly, 7| 
gallons. The British imperial gallon, used in Great Britain and 
Canada, contains 277.463 cubic inches, and a gallon of water at 
a temperature of 62 degrees Fahrenheit weighs exactly 10 pounds. 
The British imperial gallon is equal to 1.2 U. S. gallons, very 
nearly. 

What is known as the fluid ounce is the weight of yVth of a 
pint of water, or listh of a gallon of water. Since a gallon of 
water weighs S^-i pounds, a fluid ounce weighs 8| X 16 -^ 128 



= l-^j = If ounces. 



2 pints (.pt.). 
8 quarts. . . . 
4 pecks 



TABLE VI 

DRY MEASURE 



bu. 
1 



pk. 



1 = 



1 quart qt. 

1 peck pk. 

1 bushel bu. 

qt. pt. 

32 = 64 

8 = 16 

1=2 



192. The unit of dry measure is the Winchester bushel, which 
contains 2150.42 cubic inches; hence, the dry quart contains 



110 



ARITHMETIC 



§1 



2150.42 -j- 32 = 67.2 cubic inches, while the liquid quart con- 
tains 231 -^ 4 = 57 f cubic inches. A box 14 inches square and 
11.7 (say llf) deep, inside measurement, holds one bushel. The 
bushel is used in measuring charcoal and, sometimes, lime and 
coal. The British bushel is equal to 8 British imperial gallons; 
it therefore contains 277.463 X 8 = 2219.704 cu. in. and one 
British bushel equals 2219.704 h- 2150.42 = 1.03222 - Winchester 
bushels. A cylinder 18| in. in diameter and 8 in. deep holds 
exactly one bushel. 

TABLE VII 

ANGULAR MEASURE 



60 seconds (") 

60 minutes 

90 degrees 

4 quadrants (360°) 

cir. quad. (L) deg. (°) 

1 = 4 = 360 

1 = 90 

1 



1 minute ' 

1 degree ° 

1 quadrant L 

1 circle cir. 

min. (') sec. (") 

21,600 = 1,296,000 

5,400 = 324,000 

60 = 3,600 

1 = 60 



193. Table VII is very important in connection with all prob- 
lems in which the measurement of angles is necessary for their 
solution. The circle is divided into 360 equal parts called 
degrees; each degree is divided into 60 equal parts called minutes; 
and each minute is divided into 60 equal parts called seconds. 
When the circle is divided into four equal parts, each part is 
called a quadrant, and the 4 equal angles so formed are called right 
angles. A quadrant and a right angle contain 360° -;- 4 = 90°. 

194. Units of Measurement. — It should be evident from Art. 
1 that before any measurement can be made it is necessary to 

establish a unit. For linear measurements 
(Table I), the fundamental unit, from which 
the other units are derived, is the yard; the 
yard is divided into three equal parts each of 
which is called one foot; the foot is divided 
into twelve equal parts, each of which is 
called one inch. The higher units, rods, 
furlongs, and miles, are obtained by taking a 
certain number of yards. 
The units for square and cubic measure are obtained from those 

for linear measure; thus, the square inch is a square, every side of 

which measures 1 inch (see Fig. 2). 



Fig. 2. 



§1 



COMPOUND NUMBERS 



111 



mp 





















































































































































































































































































































nq 



Fig. 3. 



The square foot is a square every edge of which measures 1 
foot or 12 inches. If a square foot be divided into 12 equal parts 
by lines ah, cd, etc., Fig. 3, and again divided into 12 equal parts 
by lines mn, pq, etc., every one of the little squares that are thus 
formed will be a square inch. Between a and b, there are 12 
inch-squares, between c and d, 12 inch-squares, etc. Conse- 
quently, the foot-square has been divided 
into 12 X 12 = 144 inch-squares, and 
there are 144 square inches in 1 square 
foot. A square yard may be similarly 
divided into 9 foot-squares; hence, there 
are 9 square feet in 1 square yard. Note 
that 12 X 12 = 122 and 3 X 3 = 3^; there- 
fore, since there are 5.5 yards in a rod, a 
square rod contains 5.5^ = 30.25 = 30| 
square yards; etc. The number of square 
units in any square may be found by squaring the length of the 
side of the square. It is for this reason that the second power of 
a number is usually called the square. 

A cube measuring 1 inch on every edge is called a cubic inch 
(see Pig. 4) . A cubic foot is a cube measuring 1 foot on every 
edge. Such a cube may be divided into 12 equal layers, each of 
which measures 12 inches on each side and 1 inch high. Each 
layer may therefore be divided into 12^ equal cubes measuring 1 
inch on each edge; that is, each layer 
may be divided into 12 X 12 inch-cubes; 
and since there are 12 layers, a cube 
measuring 12 inches (==1 foot) on every 
edge may be divided into 12 X 12 X 
12 = 12^ little cubes measuring 1 inch 
on every edge. Similarly, a cubic yard 
is a cube measuring 1 yard on every 
edge and may be divided into 3^ = 27 
small cubes measuring 1 foot on every 
edge. Hence, a cubic foot contains 
12^ = 1728 cubic inches, and a cubic yard contains 3^ = 27 
cubic feet = 36^ = 46,656 cubic inches. The number of cubic 
•units in any cube may be found by cubing the length of one of 
the edges. It is for this reason that the third power of a 
number is called the cube. 

The fundamental unit of weight is the pound avoirdupois ; the 




Fig. 4. 




112 ARITHMETIC §1 

other units of weight are found by dividing or multiplying the 
pound. The fundamental unit of liquid measure in the United 
States is the United States or wine gallon, which contains 231 
cubic inches. In Canada the law allows the use of the Imperial 
gallon only for a commercial unit. Since 
the fundamental unit of dry measure is the 
Winchester bushel, which contains 2150.42 
cubic inches, the units of both liquid and 
dry measures depend upon the unit of 
linear measure. 

The fundamental unit of angular mea- 
sure is the quadrant or the right angle. 
The circle is divided into 2 equal parts by 
a line drawn through the center called a 
diameter, as ab, Fig. 5. Each half is then divided into 2 equal 
parts by another diameter cd, thus dividing the entire circle into 
4 equal parts. Each of these four equal parts is then divided 
into 90 equal parts to obtain degrees. 

TABLE VIII 

United States money 

10 mills (m.) = 1 cent ct. or jif. 

10 cents = 1 dime d. 

10 dimes (100c.) =1 dollar dol. or $ 

10 dollars =1 eagle E. 

E. $ d. ct. m. 

1 = 10 = 100 = 1000 = 10,000 

1 = 10 = 100 = 1,000 

1 = 10 = 100 

1 = 10 

Note. — With the exception of the mill and the eagle, the Canadian unit of money have 
the same names and relative values. 

196. The fundamental unit of United States money is the 
gold dollar, which is defined as 25.8 grains of gold nine-tenths 
fine; that is, the gold dollar contains 25.8 X .9 = 23.22 grains of 
pure gold. It will be noted that the scale in the table of United 
States money is 10, the same as in the Arabic system of notation; 
it is therefore called a decimal scale, and numbers may be expressed 
in different units by simply shifting the decimal point. Thus, 
$12.43 = 1243c, and 978c = $9.78. Since there are lOOj;^ in $1, 
to change dollars to cents, multiply by 100, that is, shift the 
decimal point two places to the right; and to change cents to 
dollars, shift the decimal point two places to the left. 



§1 COMPOUND NUMBERS 113 

TABLE IX 

TIME 

60 seconds (sec.) = 1 minute min. 

60 minutes = 1 hour hr. 

24 hours = 1 day da. 

7 days = 1 week wk. 

365 days =1 year yr. 

yr. da. hr. min. sec. 

1 = 365 = 8760 = 525,600 = 31,536,000 

1 = 24 = 1,440 = 86,400 

1 = 60 = 3,600 

1 = 60 

197. The fundamental unit of time, universally used, is the 
second. The year does not contain exactly 365 days, but 
365.242216 days = 365K days very nearly; hence, every fourth 
year, one day is added, making 366 days in what are called leap 
years. 

TABLE X 

MISCELLANEOUS 

12 of anything = 1 dozen doz. 

12 dozen =1 gross gr. 

12 gross =1 great gross g- gr. 



24 sheets of paper = 1 quire qr. 

20 quires (480 sheets) =1 ream rm. 

198. A gross is evidently equal to 12 X 12 = 144. It is now 
quite common practice to consider a ream as 500 sheets, except 
for stationery papers. 



THE METRIC SYSTEM 

199. The metric system is thus called because it is based on the 
meter (also spelled metre), which is equal to 39.370113 inches. 
By Act of Congress, the meter contains 39.37 inches. Sub- 
divisions of the meter are the decimeter, centimeter, and millimeter, 
and multiples of it are dekameter, hektometer, kilometer, and myria- 
meter. The scale of the system is a decimal one, the same as with 
ordinary numbers and United States money. The prefixes deci, 
centi, and milli denote respectively one-tenth, one-hundredth, 
and one-thousandth of the unit to which they are prefixed, the 
same as the dime, cent, and mill denote-one-tenth, one-hundredth, 
and one-thousandth of a dollar. The prefixes deka, hekto, kilo, 

8 



114 ARITHMETIC §1 

and myria, denote respectivly 10, 100, 1000, and 10,000 times 
the unit to which they are prefixed. 

TABLE XI 

LINEAR MEASURE 

10 millimeters (mm.) =1 centimeter cm. 

10 centimeters = 1 decimeter dm. 

10 decimeters = 1 meter m. 

10 meters = 1 dekameter Dm. 

10 dekameters = 1 hektometer Hm. 

10 hektometers = 1 kilometer Km. 

10 kilometers = 1 myriameter Mm. 

200. The fundamental unit in the foregoing table is the meter; 
the other units that are principally used are the millimeter and 
centimeter for short lengths and the kilometer for long dis- 
tances. The following table gives English equivalents for these 
units. 

1 millimeter = .03937 in. = jf s in. very nearly = .04 inch roughly. 

1 centimeter = .3937 in. = |f in. very nearly = .4 in. roughly. 

1 meter = 39.37 in. = 39f in. very nearly = 40 in. roughly. 

1 kilometer = 39,370 in, = |i mi. very nearly = f mile roughly. 

1 meter = 3.281 ft. nearly = 1.094 yd. nearly. 

It will be observed that the abbreviations for the multiples of 
the principal unit begin with a capital letter, while those for the 
sub-multiples and for the principal unit itself begin with a lower 
case letter. 

TABLE XII 

SQUARE MEASURE 

100 square millimeters (mm^).. . . = 1 square centimeter cm.^ 

100 square centimeters = 1 square decimeter dm.^ 

100 square decimeters = 1 square meter m.^ 

201. This table does not include measures of land, the principal 
unit of which is the Are = 100 m.^ The hectare, which equals 
100 ares, or 100 X 100 = 10,000 square meters, takes the place 
of the English acre, and is equivalent to 2.471 acres, say 2| 
acres, roughly. 1 square meter = 1550 square inches, very 
nearly = 10.7641 square feet, say lOf square feet, roughly. 

TABLE Xin 

CUBIC MEASURE 

1000 cubic millimeters (mm. 3).... = 1 cubic centimeter.. . c.c. or cm.' 

1000 cubic centimeters = 1 cubic decimeter. . . dm.^ 

1000 cubic decimeters =1 cubic meter m.^ 



1 cubic meter = 35.3156 cubic feet = 1.308 cubic yards. 



§1 COMPOUND NUMBERS 115 

TABLE XIV 

LIQTJID MEASURE 

10 milliliters (ml.) - 1 centiliter cl. 

10 centiliters = 1 deciliter dl. 

10 deciliters =1 liter 1- 

10 liters = 1 dekaliter Dl. 

10 dekaliters = 1 hektoliter. ... HI. 

10 hektoliters = 1 kiloliter Kl. 

202. The principal units of liquid measure are the liter and 
the hectoliter. The liter is equal in volume to 1 cubic decimeter 
= 61.0254 cubic inches = 1.0567 quarts = l^V quarts. The 
hectoliter = 26.4179 gallons - .9435 barrel. The milliliter is 
used by physicians, chemists, scientists, and others for measur- 
ing small amounts. Since a milliliter is xoVo of a hter, and since 
a liter equals a cubic decimeter = 1000 cubic centimeters, 
a miUiliter = 1 cubic centimeter, and it is customary to call 
this unit a cubic centimeter instead of a milliliter. A teaspoon 
contains }i of a fluid ounce. 1 liter = 1 dm3= 61.0254 cu. in.; 
1 quart = 57.75 cu. in. = 16 X 2 = 32 fluid ounces; H of 
a fluid ounce = 57.75 -^ 32 -^ 8 = .225586 cu. in.; 1 milliliter = 
1 cubic centimeter = 61.0254 -^ 1000 = .0610254 cu. in. 
Therefore, 1 teaspoon = .225586 -^ .0610254 = 3.6966, say 
3.7 cubic centimeters, and 1 c.c. = .27052 teaspoon, or a little 
more than }i teaspoon. Also, 1 fluid ounce = 3.6966 X 8 
= 29.5728 cubic centimeters = 30 c.c, nearly. 

TABLE XV 

MEASURES OF WEIGHTS 

10 milligrams (mg.) =1 centigram eg. 

10 centigrams =1 decigram dg. 

10 decigrams =1 gram g. 



10 grams. 



1 dekagram Dg- 



10 dekagrams =1 hektogram Hg. 

10 hektograms =1 kilogram Kg. 

1000 kilograms =1 tonne T. 

204. The fundamental unit of weight in the metric system 
is gram, which is the weight of 1 cubic centimeter of water, 
weighed under certain conditions, and is equal to 15.432 grains, 
(more accurately, 15.432356 gr.). The kilogram (frequently 
called the kilo) = 1000 grams = the weight of 1 liter of water 
= 2.2046 pounds (more accurately, 2.20462234 pounds). The 
tonne (also called metric ton) = 1000 kilograms = the weight of 



116 ARITHMETIC §1 

1 cubic meter of water = 2204.6 pounds. The metric ton is 
therefore approximately equal to the English long ton. The 
milligram is used by chemists for weighing minute quantities; 
it is equal to s\ of a grain approximately, or ^^ gr. very 
accurately. 



REDUCTION OF COMPOUND NUMBERS 

205. To reduce a compound number is to change it so that the 
denominations used to express it will be different without chang- 
ing the value of the number. For example, it may be desirable 
to express a certain number of miles, rods and feet as feet or 
inches; or it may be desired to express the number in miles and 
fraction of a mile. All such changes are called reduction or 
reducing the number to lower or higher denominations. 

206. Reducing to Lower Denominations. — The process is best 
illustrated by means of an example. Thus, express 3 mi. 126 rd. 

9 ft. in feet. Arranging the work as 

shown, multiply the number of rods in a 

mi. rd. ft. mile (320) by the number of miles in the 

3 126 9 given number, in this case 3, and the 

?^ product is the number of rods in 3 miles. 

^^^ ^^- To this add the number of rods in the 

126 

— given number, and the sum, 1086, is the 

^„i/' number of rods in 3 mi. 126 rd. Since 

-rrr there are 163^^ feet in a rod, the product 

6516 of 1086 and 16>^ = 17919 is the number 

1086 of feet in 1086 rd. To this add the 9 ft. 

17919 ft. in the given number, and the sum is 

9 ft. 17,928 ft., the number of feet in 3 mi. 236 

17928 ft. Ans. rd. 9 ft. Had it been desired to reduce 

the number to inches, the last result 

would have been multiplied by 12, and 

as there are no inches in the given number, the final result would 

have been 17928 X 12 = 215,136 in. = 3 mi. 126 rd. 9 ft. 

207. Rule. — To reduce a compound number to lower denomina- 
tions, find the product of the number representing the highest denom- 
ination in the given number and the number of units of the next 
lower denomination that will make one unit of the next higher denom- 
ination, and add to this product the number of units of that lower 
denomination, {if any) in the given number. Repeat this process 



§1 COMPOUND NUMBERS 117 

to reduce to the next lower denomination, continuing in this manner 
until the required denomination has been used. 
Example 1. — Reduce 32° 47' 19" to seconds. 

Solution. — The work is shown in the margin, and should 

be evident. Since there are 60 minutes in 1 degree, the 

1920' product of 60 and 32 is first found, to which is added the 47.' 

47 Since there are 60" in 1 minute, the product of 60 and 

1967 1967 is then found, to which is added the 19". The num- 

60 ber is now reduced to the denomination required. Note 

118020" ^^^^ ^i^^ numbers are added— minutes to minutes and 

IQ" seconds to seconds. It is not customary to write the 

1 1sn^Q" Avi abbreviations when reducing. 

Example 2. — Express 156 m., 5 cm., 2 mm. in milhmeters. 

Solution. — Write the number without abbreviations, with a period fol- 
lowing the number of units of the highest denomination, and with ciphers 
in place of any denomination that is missing. Since there are no decimeters, 
156 m. 5 cm. 2 mm. = 156.052 meters. The process is exactly the same as 
writing a number in the Arabic system of notation. To reduce 156.052 m. 
to millimeters, multiply by 1000 by shifting the decimal point 3 places to 
the right, and 156.052 m. = 156.052 mm. Ans. 

Had it been desired to express the above number in decimeters, 
multiply it by 10 by moving the decimal point 1 place to the 
right, obtaining 1560.52 dm.; to express it in centimeters, 
move the decimal point two places to the right, obtaining 
15605.2 cm. 

208. Reducing to Higher Denominations. — The process is best 
illustrated by an example. Thus, reduce 215,136 in. to higher 
denominations. 

Since there are 12 inches in 1 

215136 in. (12 foot, the number of feet in 215,136 

17928.0 ft. (16.5 inches may be found by dividing 

1^ 1086 215,136 by 12; there is no remainder, 

1428 and the quotient is 17,928 ft. There 
1320 • 

are 16.5 feet in 1 rod; hence, the 

1080 

QQ^ number of rods in 17,928 feet is 

-— ^ g A f+ found by dividing 17,928 by 16.5. 

1086 rd. (320 "^^^ quotient is 1086 rd., and the 

960 3 ^i remainder is apparently 90 ft. In 

126 rd. reality, however, it is 9 ft., because 

3 mi. 126 rd. 9 ft. Ans. the cipher in 90 is the one following 

the decimal point in the dividend, 

and has no value. Since there are 320 rods in 1 mile, divide 



118 ARITHMETIC §1 

1086 rods by 320 to find how many miles there are in 1086 rd.; 
the quotient is 3 mi., and the remainder is 126 rd. Therefore, 
215,136 in. = 3 mi. 126 rd. 9 ft. 

Note that after any division has been completed, the remainder 
is of the same denomination as the dividend; this is necessarily 
the case, since the remainder is a part of the dividend. The 
quotient, however, is of a higher denomination than the 
dividend. 

209. Suppose that instead of reducing the inches to a compound 
number the highest denomination of which, in this case, is a mile, 
it had been desired to express the number in miles and a fraction 
(decimal) of a mile. In such case, proceed in exactly the same 
manner as before, except that the quotient will in all cases be a 
mixed number (when there is a remainder), the division being 
carried to as many decimal places as is desired. Thus, in the 
example just given, 9 ft. =■ 9 ^ 16.5 = .545454+ rd. Adding 
this to the 126 rd., the sum is 126.545454+ rd. Then, 126.545454 
-r- 320 = .39545454+ mile, which added to the 3 mi. makes 
3.3954545 mi. 

210. To reduce a decimal of one denomination to units of a 
lower denomination, proceed exactly in accordance with the rule 
of Art. 207. Thus, .3954545 mi. = .3954545 X 320 = 126.54544 
rd.; .54544 X 16.5 = 8.99976, say 9 ft. The result would have 
been exactly 9 ft. if instead of expressing the rods and feet as a 
decimal of a mile they are reduced to a fraction of a mile. Thus, 

9 -^ 16.5 = j^^=U = t\ rd.; 126 + A = 126^ = ^U^ rd. 

Iff 2 ^ 320 = illl = 2%\ = .39 Ami. This last expression wUl 
reduce to 126 rd. 9 ft. exactly. 

211. Rule I. — To reduce a denominate number to higher denomi- 
nations, begin' with the lowest denomination of the given number and 
divide the number of units of that denomination by the number of 
units required to make one unit of the next higher denomination, 
forming either a common fraction {which reduce to its lowest terms) 
or carrying the quotient to any desired number of decimal places. 
Add the quotient thus obtained to the number of units in the given 
number of the same denomination as the quotient (if any), and divide 
the sum by the number of units required to make one unit of the next 
higher denomination. Proceed in this manner until the desired 
denomination is reached. 



§1 COMPOUND NUMBERS 119 

II. //, however, it is desired to reduce a given number of units of a 
lower denomination to a compound number of higher denomination, 
divide the given number by the number of units required to make one 
unit of the next higher denomination; the quotient will be of the next 
higher denomination, and the remainder {if any) will be of the same 
denomination as the dividend. If the quotient is larger than the 
number of units required to make a unit of the next higher denomina- 
tion, divide it by the number of units required to make one unit of the 
next higher denomination. Proceed in this manner nutil the highest 
denomination is reached or a quotient is obtained that is smaller 
than the number of units required to make a unit of the next higher 
denomination. 

Example. — Reduce 123,456" to a compound number; also express it in 
degrees and decimal of a degree to 5 decimal places. 
60) 1 23456" Solution. — Dividing first by 60, tbe quotient 

60)2057' + 36" ^^ 205/' and the remainder is 36". Dividing 

„.o _1_ 17' again by 60, since there are 60' in 1°, the quo- 

oAo ^'7/ OR" A tient is 34° and the remainder is 17'. Since 34° 

is less than 90°, the number of degrees necessary 

fi^^ 12*^4 '>6" 

"">> to make a quadrant, the work ceases, and 

60) 2057.6' 123,456" = 34° 17' 36". The work for reducing 

34.29333° + Ans. to degrees and decimal of a degree is evident. 
Example 2. — Express 3250615 milligrams in kilograms. 
Solution. — Beginning with the right hand figure, move the decimal point 
one place to the left for each higher denomination until the desired denomi- 
nation is reached. Thus, say milligrams, centigrams, decigrams, grams, 
dekagrams, hektograms, kilograms, placing the pencil point on the position 
occupied by the decimal point in each case, it being to the right of the 
milligrams in the beginning. When kilograms is reached, the pencil point 
will fall between 5 and 3; consequently, 5320615 mg. = 5.320615 Kg. Ans. 

This same method may be used in reducing metric numbers 
(except those for square and cubic measure) to lower denomina- 
tions. For square measure, move the decimal point two places 
each time the name of a denomination is pronounced, and for 
cubic measure, move it three places. For instance, 59308726 
mm2 = 59.308726 m^, and 607849358 c.c. = 607.849358 m^. 
Also, 78.06342 m^ = 780634.2 cm^ = 78063420 mm^, and 
9.50783 m3 = 9507830 c.c. = 9507830000 mm^, 

212. Another way of reducing the lower units of a compound 
number to a decimal or fraction of higher denomination is the 
following: To express 23 rd. 3 yd. 1 ft. 8 in. as a decimal of a 
mile, first reduce the compound number to the lowest denomi- 
nation in the given number, in this case inches, and 23 rd. 3 



120 ARITHMETIC §1 

yd. 1 ft. 8 in. = 4682 in. Referring 
to Table I, and consulting the sub- 
sidiary table, it is seen that there are 
63360 inches in a mile; hence, 4682 
in. is /aWirth of a mile. Reducing 
this fraction to a decimal, W3%%\ = 
.073895+. Therefore, 23 rd. 3 yd. 
1 ft. 8 in. = .073895 mi. 

This method is to be preferred to 
the former one, when reducing to a 
decimal or a fraction of a higher 
denomination, since it entails no 
more work, if as much, and is, in 

general, more accurate. 
4682 in. ' 



rd. yd. 
23 3 
5.5 


ft. 
1 


115 
115 




126.5 
3 




129.5 yd. 
3 




388.5 
1 




389.5 ft. 
12 




4674.0 
8 





EXAMPLES 



(1) Reduce 5 mi. 3 fur. 36 rd. 4 yd. 2 ft. 7 in. to inches. Ans. 347,863 in. 

(2) Reduce the number in (1) to miles and decimal of a mile. 

Ans. 6.490276 mi. 

(3) Reduce 13 bbl. 23 gal. 1 pt. to pints. Ans. 3,461 pt. 

(4) Reduce the number in (3) to barrels and decimal of a barrel. 

Ans. 13.73413 bbl. 

(5) Reduce 3 T. 13 cwt. 71 lb. 12 oz. to ounces. Ans. 117,948 oz. 

(6) Reduce the number in (5) to tons and decimal of a ton. 

Ans. 3.685875 T. 

(7) Reduce 14° 9' 54" to seconds. Ans. 50,994". 

(8) Express 14° 9' 54" in degrees and decimal of degree. Ans. 14.165°. 

(9) Reduce 75,906" to a compound number. Ans. 21° 5' 6". 

(10) Reduce 75,906" to degrees. Ans. 21.085°. 

(11) Reduce 50,000 in. to a compound number. 

Ans. 6 fur. 12 rd. 2 yd. 2 ft. 8 in. 



OPERATIONS WITH COMPOUND NUMBERS 

213. Addition." — Compound numbers are added in practically 
the same manner as abstract numbers. The numbers to be 
added are arranged under one another with like denominations in 
the same columns. Then, beginning with the lowest denomina- 
tion, add the numbers in that column; if the sum is greater than 
the number of units required to make a unit of the next higher 
denomination, reduce it to the next higher denomination and carry 
to the next column the number of units so found of that next 



§1 COMPOUND NUMBERS 121 

higher denomination. The process is repeated for the second and 
subsequent columns. 

Example L— Add 15° 20' 36", 21° 53' 46", 8° 49' 28", and 76° 51' 17". 

Solution. — The numbers are arranged as shown in the margin, with 
1 ^° 90' ^fi" seconds under seconds, minutes under minutes, etc. 

_ _„ .„ The sum of the second's column is 127" = 2' 7"; 

_ .„ <^„ write the 7" and carry the 2' to the next column, 

„„ _. ^_ adding the 2' to the numbers in that column. The 

sum thus obtained is 175' = 2° 55'. The 2° is carried 



rd. 


yd. 


ft. in. 


11 


4 


2 4 


34 


1 


9 


27 


3 


1 6 


17 


1 


1 2 


90 


10 


5 21 


2 fur. 10 


4i 


2 9 




5yd. 


= 1 6 



122 175 127 ^^ ^j^g j^g^^ column, making the sum 122°, and the sum 

of all the numbers is 122° 55' 7". It is not customary 

i.L^ oo / . jxiif,. to write the sums of the columns, as shown here; only 

the final results are written, unless there are fractions as in the next 

example. 

Example 2.— Add 11 rd. 4 yd. 2 ft. 4 in., 34 rd. 1 yd. 9 in., 27 rd. 3 yd. 1 ft. 

6 in., and 17 rd. 1 yd. 1 ft. 2 in. 

Solution. — The numbers are arranged as in example 1, and the sum as 

found is 2 fur. 10 rd. 4.5 yd. 2 ft. 9 
in. Since it is inconvenient to have 
a fraction in any but the lowest de- 
nomination, reduce the .5 yd. obtain- 
ing 1 ft. 6 in., which is added as 
shown, making the final sum 2 fur. 
10 rd. 5 yd. 1 ft. 3 in. 

2 fur. 10 rd. 5 yd. 1 ft. 3 in. Ans. 

Rule. — Flace the numbers to he added under each other, with like 
denominations in the same columns. Add each column, beginning 
with that of the lowest denomination. If the sum of the numbers in 
any column is greater than the number of units required to make a 
unit of the next higher denomination, reduce the sum to the next 
higher denomination before adding the next column. When the 
sum has been found, if the number of units in any denomination 
contains a fraction, reduce the fraction of a unit to lower denomina- 
tions and add to the units of lower denomination in the sum previously 
found. 

214. Subtraction.- — The operation of subtraction is practically 
the same as in subtraction of abstract numbers. Place the sub- 
trahend under the minuend, with like denominations under each 
other. If the number of units of any denomination in the sub- 
trahend is larger than the number above it, reduce one unit of 
the next higher denomination to the next lower, add it to the 
number of units of that denomination in the minuend, and then 



rd. 
34 

27 


yd. ft. 
1 
3 1 


in. 
9 
6 


6 
6rd. 


2K 2 
1 

3 yd. ~ 


3 
6 
9 in. 



122 ARITHMETIC §1 

subtract. Add 1 to the number of units in the next column of 
the subtrahend before subtracting. This process is exactly similar 
to that employed in subtracting abstract numbers. 

Example 1.— Subtract 34° 27' 17" from 90°. 

SoLTTTiON. — Placing the subtrahend under the minuend, there are no 
seconds in the minuend; hence, 1' = 60" is added to 
90° 00' 00" the minuend, and 60" - 17" = 43". Adding 1' to 27', 

34 27 17 the sum is 28', and 60' - 28' = 32'. Adding 1° to 

55° 32' 43". Ans. 34°, the sum is 35°, and 90° - 35° = 55°. The re- 
mainder, therefore, is 55° 32' 43". 

Example 2. — From 34 rd. 1 yd. 9 in. subtract 27 rd. 3 yd. 1 ft. 6 in. 
Solution. — Arranging the numbers as in example 1, 1 ft. cannot be sub- 
tracted from ft.; hence, 1 yd. = 3 ft. is 
added to the minuend, and 3 ft. — 1 ft. = 2 
ft. Adding 1 to the 3 yd. makes 4 yd.; but 
4 yd. cannot be subtracted from 1 yd. ; hence, 
1 rd. = 5}i yd. is added to the minuend, 
making 1 + 5K = 6J^ yd., and 6>^ yd. — 
Ans. 4 yd. = 2H yd. Finally, 27 rd. + 1 rd. - 
28 rd., and 34 rd. - 28 rd. = 6 rd. The 
^2 yd. is reduced to 1 ft. 6 in. and added, the sum being 3 ft. 9 in. But 3 ft. 
= 1 yd.; hence, the number of yards is increased by 1, making the final 
sum 6 rd. 3 yd. 9 in. 

Rule. — Place the subtrahend under the minuend, with like 
denominations in the same columns. Beginning with the column 
of lowest denomination, subtract the numbers in the bottom roiv from 
those above them. If the number in any column of the subtrahend 
is larger than the number above it in the minuend, reduce one unit 
of the next higher denomination to the next lower denomination, 
add it to the minuend, then subtract, and add 1 to the number in the 
column of the next higher denomination in the subtrahend. Con- 
tinue in this manner until the entire remainder has been found. 

215. Multiplication. — A compound number may be multiplied 
by an abstract number in two ways: 1st., by multiplying the 
units of each denomination separately, and reducing the products 
to higher denominations; this is advisable when the multiplier is 
a small number. 2d, reduce the compound number so that it 
will be expressed in units of one denomination, preferably, the 
lowest denomination in the compound number, then multiply, 
and reduce the product to higher denominations; this process 
is preferred when the multiplier is a large number, say greater 
than 12, or when it contains a fraction or a decimal. 



§1 COMPOUND NUMBERS 123 

Example 1.— Multiply 56 T. 13 cwt. 72 lb. by 8. 

Solution.— The product of 8 and 72 lb. is 576 lb. = 5 cwt. 76 lb. Then 

„ 13 cwt. X 8 = 104 cwt., to which is added the 

■ ' 5 cwt. carried from the first product, making 

^ 109 cwt. = 5 T. 9 cwt. Lastly, 56 T. X 8 

= 448 T., to which is added the 5 T. carried 



453 109 576 {Tom the preceding product, making 453 

453 T. 9 cwt. 76 lb. Ans. rj. ^j^^ ^^-^ product is 453 T. 9 cwt. 76 lb. 
Example 2. — Multiply 7 gal. 3 qt. 1 pt. by 53. 

Solution. — Reducing to pints, 7 gal. 3 qt. 1 pt. = 63 pt.; 63 pt. X53 
= 3339 pt. = 6 hhd. 1 bbl. 7 gal. 3 qt. 1 pt. When reducing 3339 pt. to 
higher denominations, the first result obtained is 6 hhd. 1 bbl. 7J^ gal. 1 qt. 
1 pt. Since }i gal. = 2 qt., the final result without fractions is 6 hhd. 1 bbl. 
7 gal. 3 qt. 1 pt. Ans. 

Rule I. — Begin with the loivest denomination in the given num- 
ber, and multiply the number of units of that denomination by the 
multiplier; reduce the product to the next higher denomination by 
dividing by the number of units required to make 1 of the next 
higher denomination, and add the quotient to the product of the 
number of units in the next higher denomination and the multiplier. 
Proceed in this manner until the complete product has been found. 
If a fraction occurs in connection with any product, reduce it to 
lower terms. 

n. // the multiplier is greater than 12, or if it contains a fraction 
or a decimal, reduce the multiplicand to the lowest denomination 
given, multiply, and reduce the product to higher denominations. 

Example.— Multiply 5 yd. 2 ft. 9 in. by 11.7. 

Solution. — Reducing to inches, 5 yd. 2 ft. 9 in. = 213 in.; 213 in. X 
11.7 = 2492.1 in. = 69 yd. 8.1 in., or 12 rd. 3 yd. 8.1 in. Ans. 

216. Division. — There are two cases of division: dividing a 
compound number by an abstract number, in which case, the 
quotient is a compound number; or dividing a compound number 
by a compound number of the same kind, in which case, the 
quotient is an abstract number. The simplest method of per- 
forming the division (and in most cases, the easiest) is to reduce 
the dividend to the lowest denomination in the given number, 
divide, and reduce the quotient to higher denominations. If 
the divisor is also compound, reduce it to the same denomination 
as the dividend, and divide as in division of abstract numbers, the 
quotient being abstract. 

Example 1.— What is ^th of 143° 25' 41"? 

Solution.— 143° 25' 41" = 516,341"; 516,341" -=- 7 = 73,763" = 20° 
29' 23". Ans. A somewhat easier method of performing this division 



124 ARITHMETIC §1 

(.which may be used whenever the divisor is small) is shown in the margin. 
Here proceed as in short division. Then, 143° -f- 7 = 20° + 3° remain- 
Is! iU der; 3° = 180', which added to the 25', makes 205'. 
7)143° 25' 41" 205' 4- 7 = 29' + 2' remainder; 2' = 120", which 
20° 29' 23". Ans. added to the 41", makes 161"; and 161" -^ 7 = 
23". The work may be arranged as shown or the 
additions may be performed mentally. The latter practice is not recom- 
mended, however, as it tends to increase the liability of making a mistake. 
Example 2. — How many flasks holding 2 gal. 1 qt. 1 pt. may be filled from 
a cask holding 52 gal. 1 qt.? 

Solution. — Reducing both numbers to pints, 2 gal. 1 qt. 1 pt. = 19 pt.; 
52 gal, 1 qt. = 418 pt.; 418 -r- 19 = 22; hence, 22 flasks may be filled, Ans. 

Rule. — If the divisor is abstract, reduce the dividend to the lowest 
denomination in the given number, divide, and reduce the quotient 
to higher denominations. If both dividend and divisor are com- 
pound, reduce them to the lowest denomination in either number and 
divide; the quotient will be abstract. 



EXAMPLES 



(1) Find the sum of 17 cwt. 26 lb. 9 oz., 15 cwt. 83 lb. 11 oz., 22 cwt. 55 lb. 
6 oz., 24 cwt. 71 lb. 8 oz., and 18 cwt. 48 lb. 5 oz. Ans. 4 T. 18 cwt. 85 lb. 7 oz. 

(2) What is Ath of 360°? Ans. 32° 43' 38tt". 

(3) Add 19 rd. 4 yd. 1 ft. 4 in., 31 rd. 2 yd. 9 in., 26 rd. 5 yd. 2 ft. 6 in., 
and 35 rd. 1 yd. 1 ft. 10 in. Ans. 2 fur. 33 rd. 3 yd. 5 in. 

(4) The sum of the three angles of any plane triangle is 180°; if two of 
the angles of a triangle are 36° 13' 26" and 64° 45' 40", what is the other 
angle? Ans. 79° 0' 54". 

(5) A piece of tape line 2 rd. 1 yd. 2 ft. 5 in. long was used to measure the 
distance between two points about one mile apart. The tape was applied 
172 times; what was the distance between the points? 

Ans. 1 mi. 2 fur. 2 yd. 1 ft. 8 in. 

(6) The distances between the centers of the faces of a number of pulleys 
on a main shaft were as follows: 5 ft. 8 J in., 8 ft. 6i in., 7 ft. 3| in., 10 ft. 
9| in., 9 ft. 4| in., 6 ft. 7i in. What was the distance between the first and 
last pulleys? Ans. 48 ft. 3| in. 

(7) A coal hod was found to hold 38 lb. 13 oz. of coal; how many times 
will a ton of coal fill the hod ? Ans. 51.5+times. 

(8) Bought a car of pulp weighing 47,526 lb. on which the freight rate was 
17 cents per hundredweight; how much was the bill for freight? Remember 
that all freight rates are computed on the basis of 2240 pounds to the ton, 
hence, a hundredweight here contains 112 pounds. Ans. $72.14. 

(9) The weight of a cargo of pulp was 3923 tons 1458 pounds; what was 
the freight bill, if the rate was $12.75 per ton? Ans. $50,026.55. 

(10) What was the cost of a belt 95 ft. 7M in. long at 32(^. per foot? 

Ans. $30.60. 

(11) A machine makes 1836 pounds of paper per hour; in what time will 
it make 273^ tons of 2240 pounds. Ans. 33 hr. 33 min. 



§1 THE ARITHMETICAL MEAN 125 

THE ARITHMETICAL MEAN 

217. The arithmetical mean of several numbers (or quantities) 
is the quotient obtained by dividing the sum of the numbers (or 
quantities) by the number of numbers (or quantities); it is the 
same as the average of several numbers (or quantities), and is 
usually called the mean. For example, if goods were bought from 
three different firms at discounts of 32%, 35%, and SQ}i%, the 
mean, or average, discount is (32 + 35 + 36.5) -^ 3 = 34.5%. 
This result presupposes that the same amounts were bought from 
each firm. 

Now suppose that from one firm, the bill was $126.40 and the 
discount was 32%; from another firm, the bill was $87.65 and the 
discount was 35%: from the third firm, and bill was $68.83 and 
the discount was 36.5%; what was the mean discount? Here 
it is necessary to multiply the amount of each bill by the discount 



126.40 X 32% 


(a) 
= 4044.8% 


$126.40 X 


.68 


= $ 85.95 


87.65 X 35 


= 3067.75 


87.65 X 


.65 


= 56.97 


68.83 X36.5 


= 2512.295 


68.83 X 


.635 


= 43.71 



282.88 Xx =9624.845 $186.63 

9624.845 ^ 282.88 = 34.024% = x 
$282.88 X (1 - .34024) = $186.63. 

and then divide the sum of the products by the sum of the bills. 
The result as shown herewith is a mean discount of 34.024%. 
To prove that this is correct, multiply the amount of each bill 
by 1 minus the discount and add the products; the sum will 
evidently be the total amount to be paid, and it must equal the 
total amount less the mean discount, which is shown to be true, 
both being equal to $186.63. The process is exactly the same 
as would be followed in finding the average (mean) price paid for 
126.4 lb. of some article at 32 cents per pound, 87.65 lb. at 35 
cents per pound, and 68.83 lb. at 36.5 cents per pound. The 
reason for the process is easily found. If the discount on $1 is 
32%, the discount on $126.40 is 126.4 X 32% = 4044.8%. 
The total discount on $282.88 is 9624.845%; hence, the average, 
or mean, discount is 9624.845% -^ 282.88 = 34.024% on $1. 

Example 1. — A machinist receives 52)i per hour in wages. For every 
hour he works over 8 hours a day, he receives time-and-a-half. His over- 
time for one week was: Monday, 2}i hr.; Tuesday, 2 hr.; Wednesday, 3 hr.; 
Thursday, 2 hr.; Friday, IK hr.; Saturday, 1 hr. What was his average 
rate per hour for that week? 



126 ARITHMETIC §1 

Solution. — The total number of regular hours of work was 6 X 8 = 48, 
for which he received 48 X 52 (i = $24.96. The number of hours that he 
worked overtime was 2.5+2+3+2 + 1.5+1 = 12; his rate for overtime 
was 1.5 X52(4 = 7^i; and he received for overtime work 12 X 1H = $9.36. 
The total amount that he received for the week was $24.96 + $9.36 = 
$34.32. The total number of hours worked was 48 + 12 =60. Therefore, 
his average, or mean, rate per hour for that week was $34.32 -i- 60 = $.572 
= 57.2 cents. Ans. 

Example 2. — It was desired to measure very accurately the distance 
between two punch marks. The result of measurements by five different 
persons was as follows: 10 ft. 8.1 in.; 10 ft. 8.16 in.; 10 ft. 7.97 in.; 10 ft. 
8.21 in.; 10 ft. 8.05 in. Which of these measurements is nearest to the 
mean, or average, of all the measurements? 

Solution. — The sum of all the measurements is 50 
ft. 40.49 in. The number of measurements is 5; hence, 
the mean is 50 ft. 40.49 in. -i- 5 = 10 ft. 8.098 in. which 
is very nearly equal to the first measurement, and is 
nearer that than any of the others. The probable 
correct value is 10 ft. 8.1 in. Note that the 40 in. was 
not reduced to feet, because it is to be divided by 5. 

Example 3. — The floor area of a room in which 36 clerks are employed is 
2960 square feet; what is the average number of square feet per clerk? 

Solution. — Evidently, the average number of square feet per clerk is 
2960 -^ 36 = 82f , or a space about 9 ft. square for each clerk, since 9^ = 
81. Ans. 



10 ft. 


8.1 in. 


10 


8.16 


10 


7.97 


10 


8.21 


10 


8.05 


50 40.49 


10 ft. 


8 . 098 in, 




Ans. 



EXAMPLES 



(1) The daily production of a paper machine for one week was as follows: 
97.6 T., 101.2 T., 98.5 T., 90 T., 103.1 T., 96.4 T.; what was the average 
daily production for the week ? Ans. 97.8 T. 

(2) The amount of sulphur used in making sulphite pulp was : In January, 
149,721 lb. of sulphur for 508 tons of pulp; in February, 141,176 lb. for 476 
T.; in March, 152,148 lb. for 519 T.; in April, 148,635 lb. for 493 T.; in May, 
147,204 lb. for 527 T.; in June, 153,630 lb. for 468 T.; in July, 152,582 lb. 
for 483 T.; in August, 151,796 lb. for 479 T.; in September, 154,881 lb. for 
492 T.; in October, 150,300 lb. for 512 T.; in November, 153,714 lb. for 483 
T.; in December, 149,566 lb. for 506 T. What was the average amount of 
sulphur used per ton for the year? Ans. 303.625 — lb. per T. 

(3) Sold 1250 lb. of paper at 22^ per pound; 6700 lb. at 20f5; 10,0001b. 
at 18ff; 5500 lb. at 21)4; and 15,000 lb. at 17)4. What was the average price 
received per pound. Ans. 18.5176 — ^. 



AEITHMETIC 

(PART 3) 



EXAMINATION QUESTIONS 

(1) Eeferring to example 1, Art. 164, what is the diameter 
to the nearest 64th of an inch of a circle whose area is 218 7 
'^- '^-p , . ^ , Ans. 16fi = 16H in. 

(2) Keferrmg to example 2, Art. 164, with what velocity will 
a ball of lead strike the ground if it fall from a height of 338 ft.? 

Ans. 147.45- ft. per sec. 

{6) Ihe area of a circle is proportional to the square of the 

diameter. The area of a circle whose diameter is 29| in. is 701 

sq. in.; what is the diameter of a circle whose area is 500 sq. in. 

to the nearest 64th of an inch? Ans. 25^ « in! 

(4) The price paid for a certain dye was $1.36 per ounce. This 
represented a dealer's profit of 331%, a wholesaler's profit of 
12i%, an importer's profit of 8|%, and a manufacturer's profit 
of 20%; what was the cost of manufacturing, after allowing 4 
cents for handhng, packing, etc.? Give result to the nearest 
^®^*: ^ , „, Ans. 66 cents per ounce. 

(5) A bill for rubber hose amounted to $235.40, hst price- dis- 
counts of 70, 30, 10 and 5% were allowed; (a) how much was 
actually paid to settle this bill? (6) what was the equivalent 
single discount? 

Ans. (L"^^'-''' 
^«^ A 1 r , . ^ ^^) 82.045%. 

(6) A sample of coal shows 10.83% ash; if the weight of ashes 
obtained in one day is 1632 lb., actual weight, about how many 
long tons of coal were burned? 

Ans. 6 T. 1629 lb. = 6.727 T. 

(7) A firm desires to pay a special bonus of 2% on the com- 
missions earned by its salesmen when the sales are in excess of 
a certain fixed amount; but the 2% is to be computed on the sales 
(commissions) after deducting the special bonus. What is the 
actual bonus, expressed as a per cent? Ans. 1.96+%. 

127 



128 ARITHMETIC §1 

(8) In a certain mill 35 men receive 42(/^ per hour, 64 receive 
IZi per hour, 15 men receive 87|?5 per hour, and 5 men receive 
$1.12| per hour; (a) what was the average wage per hour per 
man? (6) if they all received an increase of 12|%, what was the 
average wage per hour per man? 

I {a) 67.37^ per hr. 
I {h) 7b.7H per hr. 

(9) By making certain changes, a pulp miU increased its daily 
production from 88 tons to 95 tons. The total cost of operation 
increased 25% and the price was increased 12|%; what was (a) 
the profit per cent after the change, and (6) what was the gain 
or loss per cent in profits, if the profit when the daily production 
was 88 tons was 18%? a i ^^^ 14.65- % 

^'^' I (6) 18.6+ % loss. 

(10) How many gallons are equivalent to 9.24 cu. ft.? 

Ans. 69.12 gal. 

(11) Add 5 yd. 2 ft. 7 in., 3 yd. 1 ft. 8 in., 4 yd. 9 in. and 3 
yd. 2 ft. 10 in. Ans. 3 rd. 1 yd. 4 in. 

(12) The sum of the three angles of any plane triangle is 180°; 
if two of the angles of a certain triangle are 36° 14' 43" and 65° 
27' 13", what is the other angle? Ans. 78° 18' 4". 

(13) Express 4.807 mi. in miles and lower denominations. 

Ans. 4 mi. 6 fur. 18 rd. 1 yd. 11.52 in. 

(14) Express 75° 18' 18" as a decimal part of 360°. 

Ans. 0.2091805|. 

(15) What is (a) the weight in ounces of 57 c.c. of water? 
(6) what is the equivalent volume in cubic inches. 

Ans I ^""^ 2-^^^^+ °'- 

I (6) 3.4784+ cu. in. 

(16) Divide 26 mi. 6 fur. 22 rd. 3 yd. 2 ft. 6 in. by 15. 

Ans. 1 mi. 6 fur. 12 rd. 2 ft. 11.6 in. 

(17) Reduce (a) 1 mi. 6 fur. 12 rd. 2 ft. 11.6 in. to inches; 
(6) express this number in yards. . f (a) 113,291.6 in. 

^'''- \ (b) 3146M yd. 



SECTION 2 

ELEMENTARY APPLIED 
MATHEMATICS 

(PART 1) 



MATHEMATICAL FORMULAS 



g 



DEFINITIONS 

1. A mathematical formula, or, more simply, a formula, is an 

expression composed of ordinary arithmetical numbers and 
quantities indicated by letters, which shows at a glance what 
operations (addition, subtrac- 
tion, multiplication, division, {^ 
powers, and roots) are required 
to be performed in order to 
obtain a certain desired result. 
Roughly speaking, a formula is 
a short, concise expression of a Pj^ j 

rule, law, or principle. For 

instance, refer to Fig. 1, which represents a cylinder with a 
round hole throughout its entire length. The rule for finding 
the weight of this hollow cylinder may be stated as follows : 

Rule. — Multiply the sum of the diameters of the cylinder and 
hole hy their difference; multiply this product by the length of the 
cylinder, hy the weight in pounds of a cubic inch of the material of 
which the cylinder is composed, and by .7854, all measurements to 
he taken in inches. The final product will he the weight of the 
cylinder in pounds. 

To express this rule by a formula, 

let W = weight of cylinder in pounds; 

w = weight in pounds of a cubic inch of the material composing 
cylinder; 

1 



2 ELEMENTARY APPLIED MATHEMATICS §2 

D = diameter of cylinder in inches; 

d = diameter of hole in inches ; 

I = length of cylinder in inches; 
then W = .7854wl{D + d){D - d) 

This last expression is a formula, and it shows at a glance just 
what operations are required in connection with the quantities 
tt), I, D, d, and the number .7854 in order to find the value of W, 
the weight of the cylinder. All that is necessary in order to use 
the formula is to substitute in it the values of the quantities 
represented by the letters. Thus, suppose the diameter of the 
cylinder is 12 in., diameter of hole is 8 in., length of cylinder is 30 
in., and that it is composed of cast iron, a cubic inch of which 
weighs .2604 pounds; what is the weight of the cylinder? Here 
D — 12, d = S, I = 30, w = .2604; substituting these values for 
their corresponding letters in the formula, 

W = .7854 X .2604 X 30(12 + 8) (12 - 8) 
or, W = .7854 X .2604 X 30 X 20 X 4 = 490.8+ lb. 

2. When two or more letters in a formula are written with no 
sign of addition or subtraction between them or when a letter 
follows a number, multiplication is understood; multiplication is 
also understood when a letter precedes or follows a sign of aggre- 
gation when there is no + or — sign between or when two dif- 
ferent signs of aggregation follow each other; therefore, .7854wZ 
(D -{- d)(D — d) has the same meaning as though written .7854 
XwXlX {D -{- d) X (D - d). Obviously, the sign of multi- 
plication cannot be omitted between two numbers, since it would 
not then be possible to distinguish between the numbers. 

3. Positive and Negative Quantities. — Before going further, 
it is necessary to make a distinction between quantities that 
indicate exactly opposite directions or meanings. For example, 
suppose two men start from the same position, one walking due 
north and the other due south; they are evidently walking in 
exactly opposite directions . To indicate this fact mathematically, 
one man is said to walk in a positive direction, while the other is 
said to walk in a negative direction. Suppose that north is taken 
as the positive direction; then if one man walks 6 miles due north 
and the other walks 6 miles due south, the first man is said to 
walk -f- 6 miles, and the second man is said to walk —6 miles. 
Positive quantities are always indicated by the plus sign and 
negative quantities by the minus sign. A negative quantity is 



§2 MATHEMATICAL FORMULAS 3 

just as real as a positive quantity; it simply indicates a direction 
or meaning exactly opposite in character to that indicated by the 
positive quantity. 

A man's income may be considered as positive and his expendi- 
tures as negative. If going up hill be considered positive, going 
down hill will be negative. When the mercury in a thermometer 
is above zero, the reading is considered positive; when it is below 
zero, the reading is negative; hence, 65 degrees above zero is 
written + 65°, and 12 degrees below zero is written — 12°. If a 
push be considered positive, a pull will be negative. Many 
other instances may be cited; all that is necessary is that if any 
particular state, meaning, or direction be considered as positive, 
the state, meaning, or direction that is directly opposite in 
character will be negative. For instance, the direction in which 
the hands of a clock move, called clockwise, may be considered 
as positive; then if the hands are moved backward, or counter- 
clockwise, this direction will then be negative. 

4. In connection with arithmetical problems, it is always pos- 
sible to ascertain which of two numbers is the larger; or, if 
several numbers are under consideration, it is always possible to 
arrange them in their relative order of magnitude. But when 
quantities are represented by letters, it is seldom possible to 
arrange them in this manner, and it frequently happens that the 
subtraction of a larger quantity from a smaller may be indicated 
in a formula. The distinction between positive and negative 
quantities becomes of great importance when the quantities are 
represented by letters. For this reason, it is necessary to know 
how to add, subtract, multiply, and divide positive and negative 
quantities that are represented by letters. 

When quantities are represented by letters, they are called 
literal quantities, to distinguish them from those represented by 
figures only or figures combined with a name (concrete numbers) , 
which are called numerical quantities. The numerical value of 
any quantity is its value when expressed by figures. Thus, in 
Art. 1, the numerical value of D was 12; of d, 8; of w, .2604, etc. 

5. Coefficients and Exponents. — It was stated in Art. 2 that 
an expression like 5a means 5 X a. The number 5 which multi- 
plies a is called the coefficient of a; the coefficient of any literal 
quantity is always a multiplier of the quantity. In 33-^m, ISapq, 
3(a + 76), etc., S}^, 18, 3, etc. are the coefficients of m, apq, 
(a + 76), etc. 



4 ELEMENTARY APPLIED MATHEMATICS §2 

It is frequently convenient to represent the coefficients by 
letters; thus, the area of a ciicle is expressed by the formula 
A = irr^ in which tt represents 3.141592+ (usually taken as 
3.1416), and r represents the radius of the circle. Jn ISapq, 18o 
may be considered as the coefficient of pq, and 18ap may be 
considered as the coefficient of q. Here 18 is the numerical 
coefficient of apq, while a and ap are literal coefficients of pq and q, 
respectively. In general, however, the word coefficient refers 
only to the numerical coefficient. If no numerical coefficient is 
given, it is always understood to be 1. Thus, the coefficient 
of mn is +1, and mn = -\-lmn; the coefficient of —ay is — 1, 
and —ay=— lay. Note that when no sign is prefixed to a 
literal expression, it is always understood to be +; thus, 18 apq 
— -t- l^apq. The minus sign, however, is never omitted; hence, 
the numerical coefficient in —22>bx is —23. 

6. Exponents have the same meaning in connection with literal 
quantities that they have when used with numbers. For instance, 
an"^ = a X n X n; -4:a^¥ = -4XaXaX6X6X6; 
8xy = 8 X x^ X y^; etc. Note particularly that any quantity, 
whether numerical or literal, that has for an exponent is always 
equal to 1; thus, m" = 1, 56" = 1, etc. The reason for this will 
be explained in connection with division of literal quantities. 

Exponents may also be negative, in which case, the quantity 
affected with a negative exponent is always equal to 1 divided by 
the quantity when the exponent is the same, but positive; thus, 

x~^ = —i> — 15ay~^ = ^ — } etc. "When no sign is prefixed to 

X y 

a number or exponent, it is always understood to be +; hence, 

2 = +2, aj2 = rc+^, etc. 1 divided by a number or quantity is 

called the reciprocal of that number or quantity; thus, the 

1 . .1 

reciprocal of 2.5 is t^-^; the reciprocal of x^ is —^> etc. Therefore, 

any quantity affected with a negative exponent is equal to 
the reciprocal of that quantity affected with an equal positive 
exponent. 



OPERATIONS WITH LITERAL QUANTITIES 

7. Addition. — When several literal expressions are connected 
with one another by the signs + and — , their sum is always 
Understood. For instance, in the expression 27a^ — 54a^^ -f 



§2 MATHEMATICAL FORMULAS 5 

36at/2 - Sy^, the different parts connected by the signs + and - 
are called terms. The first term, 27 a^ is understood to be -^27 a^, 
the second term is -54.a^y, etc. The expression is equivalent to 
+27 a' + (-54a2y) + SQay^ + {Sy'). 

8. Like terms are those having the same literal quantities 
affected with the same exponents, regardless of the coefficients; 
thus, 5ay^ and 8ay^ are like terms, and -bm and llhm are also like 
terms. In the expression of Art. 7, no two of the terms are alike, 
whence they are called unlike terms. It might be thought at 
first that the second and third terms were alike, but they are not, 
since the letters do not have the same exponents. 

9. Like terms may be added, but unlike terms cannot be added 
— the addition of unlike terms can only be indicated as above. 

I. When adding literal quantities, a slightly different meaning 
is given to the term addition from that employed in arithmetic, 
because all numbers used in arithmetic are positive. Referring 
to Fig. 2, suppose a man desires to walk toward the point B; 



o 



-IS 



+ S +10 +15 



5 +s +ro +i; 



A 

Fig. 2. 



then any movement that takes him in the direction to the right 
may be called positive or +, and any movement that takes him 
in the opposite direction or toward the left will be negative or — , 
regardless of the point started from. Suppose he starts from A 
and takes 6 steps toward B; the distance advanced is then +6 
steps. If he now takes 9 steps more in the same direction, he will 
advance +9 steps. The total distance advanced toward B is 
evidently +15 steps; therefore, the sum of +6 and +9 is +15. 
This case corresponds to ordinary addition in arithmetic. 

II. Suppose, however, instead of walking toward B; he had 
walked toward C, starting from A, as before. If he takes first 
6 steps and then 9 steps toward C, his first advance is - 6 steps 
and his second advance is -9 steps, and his total advance is - 15 
steps; from which it is seen that -6 + (-9), or -6 —9, = - 15. 
That is, to find the sum of two negative quantities, add them and 
prefix the minus sign. 

III. Suppose he had walked first 6 steps toward B and then 9 
steps toward C; he would evidently stop at -3, since counting 



6 ELEMENTARY APPLIED MATHEMATICS §2 

9 steps to the left from +6 makes him pause at —3. Therefore, 
+6-9 = -3 or -9 + 6 = -3. 

IV. Suppose he had walked 6 steps toward C and then 9 steps 
toward B; he would evidently stop at +3. Therefore, —6 + 9 
= +3 or +9 - 6 = +3. 

From the foregoing, it is seen that when two numbers have the 
same sign, their sum is the sum of the two numbers prefixed by the 
common sign; but when the numbers have unlike signs, the sum is 
equal to the difference of the two numbers prefixed by the sign of the 
greater number. 

10. To add two like terms having one or more literal quantities 
in them, all that is necessary is to add their coefficients and 
prefix to the sum the sign of the greater. For instance, 
lax + 4:ax = llax; —3ax + 7 ax = 4ax; 12mn — 19mn 
= —7mn; etc. 

If there are more than two like terms to be added, find the sum 
of those having positive and those having negative coefficients 
separately, and then add the two sums; thus, 3a — 8a — a 
+ 6a — 11a + 4a = —7a, since 3a + 6a + 4a = 13a, —8a — a 
- 11a = -20a, and 13a - 20a - -7a. 

To show that the sum may be found by simply adding the 
coefficients, consider the expression, 7c + 4c = lie. Suppose c 
represents 5 inches; then 7c = 7 X 5 in. = 35 in., 4c = 4 X 5 in. 
= 20 in., and 35 in. + 20 in. = 55 in. But 7c + 4c = lie = 11 
X 5 in. = 55 in. In other words, 7 of something plus 4 of 
something is equal to 7 + 4 or 1 1 of something, and it makes no 
difference what this something is, whether it is 1 in. or 5 in. or 
105 in.; all that is necessary is that the terms be alike. 

11. Subtraction. — Subtraction may be defined as the difference 
between two quantities; it may also be defined as that quantity 
which when added to the subtrahend will give the minuend for 
the sum. Thus, the difference between 15 and 6 is 9, and 9 is the 
number that will produce 15 for the sum when it is added to 6; 
here 6 is the subtrahend and 15 is the minuend. The second 
definition is the better one to use in connection with subtraction 
of literal quantities. 

Referring to Fig. 2, call movement toward B positive or + and 
movement toward C negative or — , as before. 

I. Suppose two men to start from A ; let one man walk 9 steps 
toward B, and the other 15 steps toward B; how far are they 



§2 MATHEMATICAL FORMULAS 7 

apart? The distance between them is evidently 15 — (+9) 
= 6, since 9 + 6 = 15. 

11. Suppose they had walked toward C the same number of 
steps; then the distance between them is — 15 —( — 9) = —6, 
since -9 + (-6) = -15. 

III. Suppose one of the men had walked 9 steps toward B and 
the other 15 steps toward C; then the distance between them 
is —15 — (+9) = —24, the —24 meaning that the man at +9 
would have to take —24 steps to place him at —15. Also note 
that 9+ (-24) = -15. 

IV. Suppose that one of the men had walked 15 steps toward B 
and the other 9 steps toward C; then the distance between them is 
15 — ( — 9) = 24, that is, the man at —9 would have to walk 
+24 steps to get to +15. Also note that — 9 + 24 = +15. 

It will be observed in the foiegoing four cases that the difference 
may be found in each case by changing the sign of the subtrahend 
and proceeding as in addition. Thus, in I, the sign of the subtra- 
hend is +; changing it to — and adding, 15 — 9 = +6. In II, 
-15 + 9 = -6; in III, -15 -9 = -24; and in IV, 15 + 9 
= + 24. 

Therefore, to subtract one term from another, change the sign of 
the subtrahend and proceed as in addition. For example, 18ar 
- (7ar) = liar; 12bd - {22bd) = -lObd; Qm^ - (-Slw^) 
= +37w2; -22bV -(-346V) = + 12&V- 

12. Multiplication. — The general rule for the signs in multipli- 
cation is : If the coefficients of two terms that are multiplied have like 
signs, the sign of the coefficient in the product will be -{-;if they have 
unlike signs (one + and the other —), the sign of the coefficient in 
the product will be —. Thus, +3 X +5 = +15, and —3 X —5 
= +15; also -3 X +5 = -15, and +3 X -5 = -15. 

A little consideration will make this rule clear. Call a man's 
debts negative and his savings positive; then any tendency to 
increase the debts will be negative, and any tendency to in- 
crease the savings or to make the debts less will be positive. 
Now suppose the man saves 3 dollars each week for 5 weeks; 
at the end of 5 weeks, he will have saved 3 X 5 = 15 dollars. 
The 3 dollars and the 15 dollars are necessarily positive, re- 
presenting savings; the factor 5 is also positive, because it in- 
dicates that 3 dollars was saved 5 times. Hence, + 3 X + 5 
= + 15. In the same way, + 3 X — 5 = — 15, because this 



8 ELEMENTARY APPLIED MATHEMATICS §2 

operation indicates that 3 dollars was spent 5 times, the total 
amount spent being 15 dollars. Also, — 3 X + 5 = — 15, 
because this operation indicates that a debt of 3 dollars was in- 
creased 5 times, and an increase must be considered as positive, 
the total increase in debt being 15 dollars. Finally, — 3 X — 5 
= + 15; here a debt of 3 dollars is decreased 5 times, because, 
if + 5 indicates an increase, — 5 must indicate a decrease; hence, 
if a man's debt decreases, the value of his assets increase, that 
is, there is a positive change, which amounts to + 15 dollars. 
A similar line of reasoning may be applied to any similar case 
involving positive and negative, or opposite, changes or quantities. 

The product must contain all the literal quantities in both 
multiplicand and multiplier; thus, 66 X 9ay = 54:aby, the coeffi- 
cient of the product being equal to the product of the coefficients 
of the factors. Hence, —laH X 2ax^ = —lAa^axx^ = — 14 
a^x*. In other words, if the same literal quantity occurs in both 
factors, it will have an exponent in the product equal to the sum of 
the exponents in the two factors. In the last case, the exponent of 
a in the product is equal to 2 + 1 = 3, and the exponent of x is 
equal to 1 + 3 = 4. This is always the case, whatever the 
exponents, and whether positive or negative. Thus, ch^ 
X c-%-2 = c-^z^, since 2-4= -2, and 5-2 = 3; also, 2.1d-''^ 
X 3.4(^2 = 7.14^2-79, since 2.1 X 3.4 = 7.14, and 2 + .79 = 2.79. 

13. Division. — The general rule for signs in division is very 
similar to that for multiplication. If the dividend and divisor 
have like signs, the sign of the quotient is +; if they have unlike 
signs, the sign of the quotient is — . To find the quotient, divide 
the coefficient of the dividend by the coefficient of the divisor, 
and the result will be the coefficient of the quotient; to this annex 
the literal quantities in both dividend and divisor, but changing 
the signs of the exponents in the divisor, and then add the ex- 
ponents of like quantities. If any exponent then becomes 0, the 
value of the quantity having that exponent is 1, and it thus dis- 
appears from the quotient. For example, ASa^b^c -r- — 12a% 
= -462c, since 48 ^ -12 = -4, and a^¥ca-%-^ = a%H = Wc 
= ¥c. 

The proof for the law of signs in division is simple: As in 
arithmetic, the product of the divisor and quotient is the divi- 
dend (no remainder being considered) . Letting d represent the 
dividend, p, the divisior, and q the quotient, d ^ p X q. If cZ 
and p are both positive, q will also be positive, because -|- p 



§2 MATHEMATICAL FORMULAS 9 

X -\- q = -\- pq = -{- d; also, if d and p are both negative, q 
will be positive, because — p X -\- q= - pq = - d. Further, 
if d is positive and p negative, q is negative, because -pX-q 
= -\-pq = + d; also if d is negative and p positive, q is negative, 
because +pX-q=-pq=-d. Therefore, if the signs of 
the dividend and divisor are aHke, the quotient is positive; if 
they are unHke, the quotient is negative. 

In practice, the exponents of the literal quantities in the divisor 
are subtracted from the exponents of like quantities in the divi- 
dend; this produces the same result as changing the signs of the 
exponents. Thus, in the last example, the exponent of a in the 
quotient is 2-2 = 0; of &, 3- 1 = 2; and as c does not occur 
in the divisor, it goes into the quotient unchanged. 

To prove that any number having for an exponent is equal to 
1, let ?i represent any number or quantity, then n -^ n = 1; but 
n -i- n = n^~^ = n*\ Since the only way that can be obtained 
for an exponent is to divide a number or quantity by itself, it 
follows that any number or quantity having for an exponent is 
equal to 1. Similarly, n^ -^ n^ = n^~^ = w" = 1. 

14. Operations with Pol3momials. — An expression consisting 
of but one term is called a monomial (the prefix mo is a contrac- 
tion of mon meaning one); an expression consisting of but two 
terms is called a binomial (hi means two); an expression 
consisting of more than two terms is called a polynomial {poly 
means many). 

Before performing anyof the operations of addition, subtraction, 
multiplication, or division, it is always advisable to arrange the 
polynomials according to the descending powers of one of the 
letters; this is done by writing first that term containing the 
highest power of the letter chosen, then the term containing the 
next highest power, etc. until all the terms have been written. 
For example, arrange 2x^y - 2axy^ + a;^ - ^/^ + x^y^ — aH^ 
according to descending powers of x. The term containing the 
highest power of x is x^, that containing the next highest power of 
X is —aH'^, etc. consequently, the polynomial arranged according 
to descending powers of x 1% x^ - aH^ + 2x^y + x'^y'^ - 2axy^ 
— 1/^. If it were desired to arrange the polynomial according to 
descending powers of y, the result would then be —y^- 2axif 
+ x^y^ + 2xhj — aH'^ + x^. Note that in the first polynomial 
as arranged, the last term does not contain x, and in the second 
arrangement, the last two terms do not include y. 



10 ELEMENTARY APPLIED MATHEMATICS §2 

15. To add two or more polynomials, first arrange the one 
containing the most terms acording to the descending powers of 
one of the letters; then place under this the other polynomials, 
with like terms in the same columns, and add each column 
separately as in addition of monomials. 

Example 1. — Find the sum of a;^ — 3xy + y^ + x + y — 1, 2x^ + 4:xy 

— 3?/2 - 2x - 2y + 3, 3x^ - 5xy - Ay^ + 3x + 4:y - 2, and Qx^ + lOxy 
+ 5y^ + X + y. 

Solution. — Arrange the first polynomial according to descending powers 
of X, and then place under it the other polynomials with like terms in the 
same columns. When adding polynomials, it is customary to begin at the 
left, instead of at the right, as in arithmetic, and this can be done because 
there is never an5rthing to carry from one column to the next. 
The sum of the coefficients in the first column is 12, and the first term of 
the sum is 12x^. 

x^ — 3xy + X -{- y^ + y — 1 
2x2 + 4a.y _2x -3y^ -2y + 3 
3x^ - 5xy + 3x - Ay^ + 4:y - 2 
6x^ + lOxy + X + 5y^ + y 
12a;2 + 6xy + 3x — y^ + 4y. Ans. 

The sum of the coefficients in the second column is 10 + 4—5 

— 3 = 6, and the second term of the sum is Qxy. The sum of the coeffi- 
cients in the third column is 1+3 + 1— 2=3, and the third term of the 
sum is 3x. The sum of the coefficients in the fourth column is 5 + 1 

— 4 — 3 = — 1, and the fourth term of the sum is —y^. The sum of the 
coefficients in the fifth column is 1+4 + 1— 2=4, and the fifth term 
of the sum is 4y. The sum of the numbers in the sixth column is 3 — 2 

— 1=0, and the entire sum is 12^^ + 6xy + 3x — y^ + Ay. 

Example 2. — Find the sum of 4a — 5& + 3c — 2d, a + & — 4c + 5d, 3a 

— 7& + 6c + Ad, and a + 4& - c - 7d. 

4a — 55 + 3c — 2d Solution. — Since the letters in all these polynom- 

a + 6 — 4c + 5d ials have the same exponents, 1, arrange them accord- 

3a — 76 + 6c + 4d ing to the order of their letters, i.e., alphabetically, 

a + 46 — c — 7d The sum is then found in the same manner as in 

9a - 76 + 4c. Ans. Example 1. 

16. To subtract one polynomial from another, arrange the 
minuend according to descending powers of one of the letters, 
place the like terms of the subtrahend under it in the same columns, 
and subtract each term of the subtrahend from the term above it 
in the minuend, as in subtraction of monomials. If the subtra- 
hend contains a term not in the minuend, change its sign and 
write it in the remainder; and if the minuend contains a term not 
in the subtrahend, write it in the remainder with its sign 
unchanged. 



§2 MATHEMATICAL FORMULAS 11 

Example 1. — From 4x^ - 2x^ + 3a;^ - 1 + 7a; subtract 6a; + 1 - a;* 
+ 2x3 - 2x\ 
Solution. — Arranging the minuend according to descending powers of x, 

and writing the subtrahend under it with 

like terms in the same columns, begin at 

3a;* + Ax^ — 2x2 + 7a; — 1 the left and subtract each term of the 

—X* + 2x^ — 2x2 _j_ Q-j. _(_ 1 subtrahend from that above it in the 

4x* + 2x3 ^ X - 2. ^TOS.niinuend. Since 3 - ( - 1) = 4, the 

first term in the remainder (difference) 
is 4x*. The second term of the differ- 
ence is evidently 2x^, the third term is 0, the fourth term is x, and the coef- 
ficient of the fifth term is -l-(+l) = -l-l = -2. The final 
result obtained for the difference is 4x* + 2x^ -\- x — 2. 

Example 2.— From Sam^ — IZa^m^ + ISa^m^ - 29 subtract Saw^ 
+ 4a2m* + 14a*TO2 - 24:a^m - 38. 

Solution. — The arrangement according to descending powers of m, and 
allowing for missing powers in both subtrahend and minuend is shown below. 
8am5 - IZa^m^ + ISo^m^ - 29 

SawS + 4a2m* + 14a*w2 - 24a% - 38 

- 4a2m4 - IZa^m^ + Aa^m^ + 24:a^m + 9. Ans. 
As there is no term in the minuend over the second term of the subtrahend, 
change its sign and write it in the difference. There is no term in the 
subtrahend under the second term of the minuend; hence, bring it down into 
the difference as it stands. There is no term in the minuend over the 
fourth term of the subtrahend; hence, change its sign and bring it down into 
the difference. The rest of the work is evident, and the difference sought 
is -4a2m4 - IZa^m^ + 4a*m^ +24:a^m + 9. That the result as obtained 
is correct may be proved by adding the difference to the subtrahend, using 
the work as it stands; the sum is the minuend. 

17. To multiply a polynomial by a monomial, simply multiply 
separately each term of the polynomial by the multiplier. It is 
customary to arrange the multiplicand according to the descend- 
ing powers of one of the letters and begin the multiplication with 
the term containing the highest power, i.e., begin at the left. 

Example. — Multiply 5x^ — 2ax^ + Ta^x — 14a3 by AaH^. 

Solution. — The multiplicand is already arranged according to the 

descending powers of x. Multiplying each term separately by Aa^x^, the 

5x3 _ 2ax2 + 7a^x - Ua^ product is 2Qa^x^ - SaH^ + 28 

4(i2^3 a*a;* — SGo^x^. The product is 

20a2x« - 8a3x« + 28a%* - 56a«x3. Am. ^^^^^y fo^^d in this case without 

placmg the multiplier under the 
multiplicand, but this has been done in the margin to show how the work is 
arranged. Beginning at the left and taking each term of the multiplicand 
in succession, 5x3 x 40^x3 = 2QaH^, — 2ax^ X "^aH^ = - Sa^x^, etc. 

18. To multiply a polynomial by a binomial or another polyno- 
mial, arrange both multiplicand and multiplier according to 



12 ELEMENTARY APPLIED MATHEMATICS §2 

descending powers of one of the letters, placing the first term of 
the multiplier under the first term of the multiplicand. Then 
multiply each term of the multiplicand by the first term of the 
multiplier; the result will be the first partial product. Multiply 
each term of the multiplicand by the second term of the multiplier; 
the result will be the second partial product, which is written 
under the first partial product with like terms under each other. 
Proceed in this manner until all the terms of the multiplier have 
been used. Now add the partial products, and the sum will be 
the entire product. 

Example 1.— Multiply 4a^ - Sab + &b^ by a - 36. 

Solution. — Multiplying the multiplicand by a, the product is ^a^ 

— 5a^b + 6ab^, which is written under 

the multiplier. Now multiplying by 

4q2 _ 5^^ + 6&'' ~ ^^' *^^ product'of 4a2 and — 36 is 

a — Sb ~ 120^6, which is written in the same 

. 3 _ — _ 21 I ^~^ column as the — 5a% of the first par- 

ir, ,1. I ir 19 -I07., tial product. Then — 5ab X — 36 

- 12a26 + 15a62 - 186' , _\„ , • , • •,, • ,, , 

= loab^, which is written in the col- 



4a3 - 17a'b + 21ab^ - 186". Ans. ^mn containing 606^. 66^ X - 36 

= — 186^, which is written to the 
right of the preceding product. Add- 
ing the two partial products, the entire product is 4a^ — 170^6 + 21ab^ — 186'. 

Example 2. — Multiply D + dhy D — d. 

Solution. — The work is shown in the margin, and should be evident 
without any special explanation. The entire pro- 
duct is D^ — d^. It is to be noted that D and 
D + d ^ are to be treated as though they were two different 

D — d letters. This result shows that the product of the 

D^ -{- Dd sum and difference of two numbers or quantities is 

— Dd — d^ equal to the difference of their squares, and the for- 

2)2 — d^. Ans. niula in Art. 1 may be written ^ = .7854w;Z (Z)2 — d^). 

In the example that is given in connection 
with this formula, Z> = 12 and d = 8; D"^ = W- 
= 144 and ^2 = 8^ = 64; 144 - 64 = 80. But D 
+ d = 12 + 8 = 20; D - d = 12 - 8 = 4; {D -\- d) (D - d) = 20 X 4 
= 80, the same result in either case. 

Example 3.— Multiply x*+ 2x^ + x^ - Ax - U by 5x^ - 2x + 3. 
Solution. — The work is shown below and requires no special explanation. 
x^ + 2x^ + x^ - 4a; - 11 
5x^ - 2a; + 3 
5x6 + 10x5 + 5a;* - 20^' - 55x^ 

- 2x5 - 4x* - 2x' + 8x2 ^ 22x 

3x* + 6x3 4, 3a;2 _ i2x - 33 

5x6 ^ 8x5 + 4a;4 _ i5a;3 _ 443.2 _|. jqx - 33. Ans. 



§2 MATHEMATICAL FORMULAS 13 

It will be observed here that when the multiplicand and multiplier are 

arranged according to the descending powers of some letter, each partial 

product begins one place farther to the right than the preceding partial 

product, and that all terms of the partial products follow in regular order. 

Example 4. — Suppose the multiplier in the last example had been x^ 

— 2x +3; what would the product have been? 

Solution. — The work is shown herewith. This example has been selected 

X* + 2x3 _^ a;2 - 4x - 11 

x^ -2x +3 

a;8 + 2x5 + X* - 4x3 _ iia;2 

- 2x5 _ 4a;4 _ 2x^ + 8x2 ^ 22x 

3x4 + 6x3 4, 3a;2 _ i2x - 33 
x« + lOx - 33. Ans. 

to show how terms will sometimes disappear in multiplication, the product 
in this case being x^ + lOx — 33, the terms containing x^, x\ x^, and x^ 
having disappeared. 

19. To divide a polynomial by a monomial, arrange the polyno- 
mial according to the descending powers of one of the letters and 
divide each term separately by the divisor. Thus, to divide 
21a4&2 _ I2a%^ + 24:a%^ - d3a¥ by dah^, arrange the work in 
exactly the same manner as for short division in arithmetic. 

Sa¥) 21a^b'~ - Ua'^b^ + 24:a^¥ - 33ob^ 

7a3 - Aa^b + 8a¥ -lib'. Ans. 
Beginning with the first term of the dividend and dividing it by 
3ab^, the quotient is 7a^, which is the first term of the quotient 
sought. The other terms are found in the same way, the quotient 
being 7a^ - 4:a^b + Sah^ - llbK 

20. To divide a polynomial by a binomial or by a polynomial, 

arrange the dividend and divisor according to descending powers 
of one of the letters, placing them in the same relative positions 
as for long division in arithmetic. Divide the first term of the 
dividend by the first term of the divisor, and the result will be the 
first term of the quotient; multiply the divisor by the first term 
of the quotient and subtract the product from the dividend. 
Divide the first term of the remainder just found by the first 
term of the divisor, and the result will be the second term of the 
quotient, which multiply into the divisor and subtract the prod- 
uct from the remainder first found. Proceed in this manner until 
a remainder of is obtained or until the first term of the last re- 
mainder is of loioer degree than the first term of the divisor, in 
which case, write the remainder as the numerator of a fraction 
whose denominator is the divisor. 



14 ELEMENTARY APPLIED MATHEMATICS §2 

Example 1. — Divide a^ — 2ab^ + b^ hy a — b. 

Solution. — Arranging the dividend and divisor according to the descend- 
ing powers of a, place the divisor to the right of the dividend, with a 

curved hne between, and draw a hne 

flS _ 2ab^ + &' (a — b under the divisor to separate it from 

o3 _ 025 a^ _^ ab - 62. Ans. *^^ quotient. Then, a^ -i- a = a\ 

~ iK _ 2 ?)2 -1- 63 (A') *^^ ^^^^ term of the quotient. 

27 _ ,2 Multiplying the divisor by a* and 

— subtracting the product from the div- 

— ao -t K ) idend, the remainder is shown at (^), 

JZ - — which is arranged according to de- 

(C*) scending powers of a. Then, a^b 

-ir a = ab, the second term of the 
quotient. Multiplying the divisor by ab and subtracting the product from the 
remainder (^), the new remainder is shown at (5). Then, —ab^ -^ a = —b^, 
the third term of the quotient. Multiplying the divisor by —6* and sub- 
tracting the product from the remainder at (B), the difference is 0, showing 
that the division is exact. The quotient sought is a^ + ab — b^. 

Example 2.— Divide 64^5 - 486aS by 2x - 3a. 

Solution. — The work is shown below and requires no special explanation. 
64a;6 - 486o5 ( 2x - 3a 

64a:5 - 96ax ^ 32a;4 + 48ax^ + 720^x2 + lOSa^x + 162a*. Ans. 
96ax* - 486a5 
96aa;4 - 144a2a;3 



144a2a;3 - 486a5 
144a2a;3 - 21Qa^x^ 



21Qa^x^ - 486a« 

21&a^x^ - 324:a^x 

324a4x - 486as 
324a*x - 486a'' 

That the quotient as obtained is correct may be proved by multi- 
plying it by the divisor; the product will be the dividend. 

Example 3.— Divide 9n^ + llcn^ + 17c^n by 3n^ + 4c. 
Solution. — To obtain the second term of the quotient, note that 
9n* + llcn2 + 17c^n (3n^ + 4c 

^!^1+12^ 3n2-|c + §f^i^. Ans. 

- cn^ + 17c^n 9«== + 12c 

— cre^ — |c^ 

17c^n + fc" ^ Slc^r; + 4c2 ^ c'CSln + 4) 
3to2 +4c ~ 9^2 + 12c ~ 9tc2 + 12c 
— cn^ -T- 3n^ = — ic. Note further that the exponent of n in the first 
term of the second remainder, 170%, is smaller than the exponent of n in 
the divisor; hence, the remainder is said to be of lower degree than the 
divisor, and the division ceases, the remainder being written as the numerator 
of a fraction whose denominator is the divisor. The numerator contains the 
fraction | ; to get rid of it, multiply both numerator and denominator of the 



§2 MATHEMATICAL FORMULAS 15 

entire fraction by 3 (as shown in arithmetic, this does not alter the value of 
the fraction), and the result is -q-YaTTo"' ^* ^^ readily seen that Slc^n 
-f- 4c2 = c^{51n + 4), since if the parenthesis be removed and the terms 
within it are multiplied by c^, the product will be the binomial Slc^w + 4c2. 

c^(51n + 4) 
Therefore, the quotient is Sn* — |c H — q „ , . 2^ ' 

21. Signs of Aggregation.' — The signs of aggregation are used 
much more freely in connection with literal quantities than with 
numerical quantities, the reason being that numerical quantities 
can be readily combined to form a single number, which is not 
possible with literal quantities, unless the terms are alike. The 
rule given in arithmetic for removing the signs of aggregation 
applies to both numerical and literal quantities. Thus, a + 
[b + (c— d)] = a -}- [b + c — d] = a -\- h -\- c — d. Here, as in 
arithmetic, when one sign of aggregation includes another, the 
inner one is removed first. This is particularly advisable when 
some of the signs are negative. Thus, a — [b — {c — d)] 
= a — [b — c-{-d] = a — b + c — d. 

Example 1. — Remove the signs of aggregation from 2x — {Sy — [4a; 

- (52/ -6a;)]}. 

Solution. — 2x — {3y — [4a; — (5y — 6a;)]} = 2x — {3y — [Ax - by 
+ 6x]} =2x - {3y - [lOx - by]] = 2x - {3y - 10a; + 5y} = 2x - {Sy 

— lOxj = 2x — Sy + lOx = 12a; — 8?/. First remove the parenthesis; since 
the quantities enclosed by it are subtracted from 4a;, their signs must be 
changed, thus making the expression within the brackets 4a; — 5y + 6a;, 
which is equal to lOx — by. Next remove the brackets, obtaining for the 
expression within the brace 3?/ — lOx + by, which is equal to Sy — lOx. 

Now removing the brace, the original expression has reduced to 2x — 8y 

+ 10a;, which is equal to 12a; — 8y. Ans. 

Example 2. — Remove the signs of aggregation from 
bm{{a -b)[a^ - 46(a2 - b^)]} 

Solution.— 5m{ (a - b)[a^ - 4:b{a^ - b^)]} = bm{ (a - b)[a^ - ia% 
+ 463]} = bm{a^ - 4:a^b + Aab^ - a^b + Aa^'b^ - 46^ = bma^ - 20ma% 
~bma% + 20ma%^ + 20mab^ — 20mb\ Ans. Having removed the paren- 
thesis, the polynomial within the brackets is multiplied by the binomial, 
a — b, and the brackets are removed. Now removing the brace and 
multiplying the last product by bm, clears the expression of all the signs of 
aggregation. 

22. When two or more terms of an expression have a common 
factor, the terms may be included in a parenthesis or other sign 
of aggregation and the common factor placed outside as a multi- 
plier. For instance, the polynomial x^ + 2ax + a^ may be 
written x{x -\- 2a) -|- a^ or a;^ + a (2a; + a), if desired. When the 



16 ELEMENTARY APPLIED MATHEMATICS §2 

first term within the parenthesis has the minus sign before it, as in 
x^ — 2ax -{- a^ = x^ -\- a{ — 2x -\- a), it is desirable to have the 
first term positive, and this may be done in the present case by 
changing the order of terms, writing the expression x^ -\- a(a 

— 2x) . If, however, it is not desired to change the order of terms 
or if both terms are negative, place the minus sign outside the 
parenthesis and change the signs of all the terms within it; in such 
case, the last expression becomes x^ — a(2x — a). That this is 
correct will be clear when it is noted that the expression as it 
stands means that a(2x — a) is to be subtracted from x^, and in 
subtraction, the signs of the subtrahend are changed, thus making 
the expression x^ — 2ax + o,^, the original form when the paren- 
thesis is removed. 

Suppose it were desired to divide x^ — x^y + xy^ — y^ by 
x — y. This may be done in the regular way, but a somewhat 
easier method in this case is the following: x^ — xhj + xy^ —y^ 
= x^(x — y) -\- y^{x — y) = (x — y){x^ + y^), and this last ex- 
pression divided by x — y gives x^ + y^ for the quotient. That 
x^(x — y) -{- y^(x — y) = (x — y)(x^ + y^) is evident, because a; — y 
is a factor common to both terms, and if the parenthesis be 
removed from {x^ + y^), the expression reduces to the preceding 
one. 

If more than one of the signs of aggregation are used, the inner 
one is used first and' the terms enclosed by it are treated as a 
single term when employing the next sign. For example, 3m^ 

— 12m^n — 6mn^ — 18n^ = 3m^ — 12m^n — Qn^(m + Sri) =3m^ 

— 6n[2m^ + n(m + 3w)] = 3{m^ — 2n[2m^ + n(m + 3n)]}. 
Note that the signs of the terms within the parenthesis are not 
changed when the brackets are written, the entire expression, 
n(m + 3rz), being treated as though it were a single letter. 

23. The Signs of a Fraction.^ — A fraction has three signs, one 
for the numerator, one for the denominator, and one for the 
entire fraction, the latter showing whether the fraction is to be 
added or subtracted. Thus, let a be the numerator and b the 
denominator, then the fraction may have one of the following 
. , -\-a a , —a a , -\-a a -{-a a 

forms: +^ = ^^ + +5 = "6^ + ^ = "6' " +6 = "6' 

— a a -\-a _a —a a 

~ +6"'6' ~ '-b' ^b' ~ ^h^ ~b- 

To prove that these results are correct, apply the law that if 
any quantity be multiplied and divided by the same number or 



§2 MATHEMATICAL FORMULAS • 17 

quantity its value is unchanged; thus, 3 X -1 = — 3, and —3 
4- — 1 =3. Now remembering that to divide a fraction, its 
numerator may be divided or its denominator multipHed by the 
/ —a\ . — a ^ — l __«_-, 

divisor, -IX [+^) -^ -I = ip^ - 5, -L 



= + — —,-, = +^=y: etc. Now since multiplying by 

' +0 0' 

— 1 changes the sign of the fraction and dividing by — 1 changes 
the sign of the numerator or denominator, as the case may be, it 
follows that if two of the three signs of the fraction are changed, the 
value of the fraction is not altered; hut, if one sign only or all three are 
changed, the value is altered. 

Suppose the fraction has the form — ^2_a^ — ) *o ^^^^ *^^ 
sign before the first term of the numerator + , change the sign 
of each term of the numerator and the sign of the fraction, 

obtaining - ^ o f' - The same result might have been ob- 

^ x^ — a^ 

tained by changing the signs of both numerator and denominator, 

obtaining ^ ?" !^ , but this changes the order of the terms in the 
^ a^ — x^ 

denominator, which is not usually desirable. 

-- + 24c^ -c + 48c^ _ c-48c^ 

Similarly, 4^2 + 7^ _ c " Sn^ + 14w - 2c " Sn^ + 14n - 2c* 
Here, to get rid of the fraction in the numerator, both numer- 
ator and denominator are multiplied by 2 (this does not alter 
the value of the fraction); then the sign before the fraction and 
the signs of the terms in the numerator are changed. If the 
signs of the numerator and denominator are changed, instead of 
the sign of the numerator and the sign of the fraction, the result 

•u u c-48c^ _ c-48c^ 

will be _g^2 _ j4^ + 2c ~ 2c-14n-8n2- 



EXAMPLES 



(1) Find the sum of a^ - b'' + Sa^b - bah\ Za^ + 36^ - Zah^ - ^a% 
a.3 + 63 4. ^a% 2a^ - 46' - 5ab% Qa^b - Sa^ + lOob^, and 2b3 - Ta^ft 
+ 4ob2. Ans. 4a2 + a% + ah^ +h\ 

(2) From Sm' - 2m^ - m -1 subtract 2m^ - Sm^ + m + 1. 

Ans. m^ + m^ — 2m — 8. 



2 



18 ELEMENTARY APPLIED MATHEMATICS §2 

(3) Add a^ - Say + y^ + a + y - 1, Aay - 2a + 2a'' - 3y^ - 2y + S, 
3a2 - 5ay - 4y^ + 4?/ - 2, and 5y^ + lOay + 6a^ + y + a. 

Ans. 12o2 + Gay - y^ + Ay. 

(4) Subtract a - h - 2{c - d) from 2(a - 6) - c + d. 

Ans. a — h -\- c — d. 

(5) Subtract a^ + a;^ — ax from Za^ — 2ax + x^. Ans. 2a^ — ax. 

(6) Multiply 2r3 + 4r-2 + 8r + 16 by 3r - 6. 4ns. 6r« - 96. 

(7) Multiply 4m2 — 3mn — n^ by 3m - 2n. 

Ans. 12m3 - 17m^n + 3mn^ + 2nK 

(8) Divide a^ - 6a^ + a^ + 4 by a^ - 1. Arts, a^ - 5a^ - 4. 

(9) Divide 32s5 + t^ by 2s + t. Ans. 16s^ - Ss^i + 4tsH^ - 2st^ + t\ 

(10) Remove the signs of aggregation from 

7a - {3a - 2[4a - 3(5a - 2)]}. Ans. 12 - 18a. 

(11) Remove the signs of aggregation from 

15 - 2y{x +y) {x- 2y) - Ay[2x - 3y(3 - x)] 

Ans. 16 — Sxy + 3Qy^ — lOxy'^ — 2xhj + 4?/^. 

(12) Eaclose within parenthesis the third and fourth terms, also the fifth 
and sixth terms, and then enclose all except the first term in brackets, of 
the expression 1 — 20w — 36TOn + \2n^ + Sm^n + Sre^. 

Ans. 1 - 4n[5 + 3(3m - n^) - 2{m^ + n)]. 

— 7 . 

(13) Change the signs of the fraction — -r^ without changing its value. 

, 7 

Ans. — 



c + 1 
(14) Change the signs of the fraction 2 _ o — ZTi without changing its 

, , 2a - 8 2a - 8 

value. Ans. ; ^ r or, . , _ -.■ 

a^ — 2a — 4 ' 4 + 2a — a^ 



EQUATIONS 

24. An equation is an expression of equality between two sets of 
quantities; thus 4^ = 64, (a + 6)^ = o^ + 2ab + b^, and a^ 
+ 36 = 2a — 7 are equations. It will be noted that an equation 
consists of two parts, or members, which are separated by the sign 
of equality. The part on the left of the sign of equality is called 
the first member, and the part on the right is called the second 
member. In the first two equations given above, the second 
member is merely another way of writing the first member, the 
second member being obtained by performing the operations 
indicated in the first member. Equations of this kind are not 
true equations; they are called identical equations. The third 
equation is a true equation, and since the second member cannot 
be obtained by rewriting or performing any indicated operations 
in the first member, it is called an independent equation. Every 
formula such as was given in Art. 1, is an independent equation. 



§2 MATHEMATICAL FORMULAS 19 

It is to be noted that in every equation, the two members are 
exactly equal, in the same sense that 2 = 2 or 3 + 4 = 7. 
Bearing this in mind, the following law will at once be evident: 

If the same quantity he added to or subtracted from both members, 
if both members be multiplied or both divided by the same quantity, 
or if both members be raised to the same power or the same root of 
both members be taken, the equality of the members is unaltered. 
For instance, take the identical equation 16 = 16; adding 4 to 
both members, 16 + 4 = 16 + 4, or 20 = 20; if 4 be-subtracted 
from both members, 16 - 4 = 16 - 4, or 12 = 12; if both 
members be multiplied by 4, 16 X 4 = 16 X 4, or 64 = 64; if 
both members be divided by 4, 16 -^ 4 = 16 -^ 4, or 4 = 4; if 
both members be squared, 16^ = 16^, or 256 = 256^ finally, if 
the square root of both members be taken -x/lQ = \/l6, or 4 = 4, 
What is true of an identical equation, insofar as the above law is 
concerned, is also true of any independent equation. For ex- 
ample, if x^ -{- 5x = 16 and 4 be subtracted from both members, 
the equation becomes x^ -\- 5x - 4: = 16 - 4 = 12; and if 16 be 
subtracted, the equation becomes x^ -{- 5x — IQ = 0, 

25. Transformation of Equations. — To transform an equation 
is to make alterations in the members without destroying their 
equality. This is done by applying the law of Art. 24. 

Case I. — A term may be transferred from one member to the 
other by changing its sign; this is called transposition. Consider 
the equation 40 - 6a; -16= 120 - 14rc. Performing the opera- 
tions indicated in the first member, 24 - 6a; = 120 - 14a;; this 
operation is called collecting terms. Transposing the 14a; to the 
first member and the 24 to the second member, 14a; - 6a; = 120 
- 24. Collecting terms, 8a; = 96. Dividing both members by 
S, X = 12. To prove that x = 12, substitute 12 for x in the 
original equation; if the two members are then equal, 8 is the 
correct value for x. Thus, 40 - 6 X 12 - 16 = 120 - 14 X 12, 
or 40 - 72 - 16 = 120 - 168, whence, - 48 = - 48, and the 
equation is said to be satisfied. 

To prove the rule for transposition, consider the second equa- 
tion above, 24 - 6.1; = 120 - 14a;. If 14a; be added to both 
members (which, by the law of equations, does not destroy the 
equality), the equation becomes 14a; + 24 - 6a; = 120 - 14a; 
+ 14a; = 120. Note that as the result of this operation, the term 
14a; has here been transferred (transposed) to the first member 



20 ELEMENTARY APPLIED MATHEMATICS §2 

and that its sign has been changed from — to +. Subtracting 
24 from both members of the equation 14a: + 24 — 6a; = 120, 
the equation becomes, 14a; + 24 — 24 — 6a; = 120 — 24, or 
14a; — 6a; = 96, from which, 8a; — 96. Note that as the result 
of this operation, the term 24 has been transposed to the 
second member and that its sign has been changed from + to — . 
Evidently, therefore, a term may be transposed from one member 
to the other by changing its sign. Having reduced the equation 
to 8a; = 96, both members may be divided by 8 without 
destroying the equaHty, according to the law of equations. 

X CO ^ 

Case II. — Find the value of a; in ^ ~ q ■^~ ^ ^ ^'^' '^^® ^^^* 

step is to clear the equation of fraction, which may be done by 
multiplying both members by each denominator in succession. 

Thus, multiplying first by 2, a; ^ ~^ 'k ~ ^^' multiplying 

6a; 
by 3, 3a; — 2a; H — ^ = 264; multiplying by 5, after first com- 
bining 3a; and — 2.a;, 5a; + 6a; = 1320; dividing both members 
by 11, since 5a; + 6a; = 11a;, a; = 120. Substituting this value 

f • +V, • • 1 ,.120 120 , 120 . . , 
of X m the origmal equation, -^ ^ — I — r' =44; whence, 

60 — 40 + 24 = 44, and the equation is satisfied. 

Multiplying by 2, 3, and 5 is the same as multiplying by 2 X 3 
X 5 = 30. Therefore, instead of multiplying by these numbers 
separately, both members of the original equation might have 

30 X a; 

been multiplied by 30, and the result would have been — ^ — 

30 V r 30 V r 
^+ ^-f-^ = 30 X 44; whence, 15a; - 10a; + 6a; = 1320, 

1320 
or 11a; = 1320, and x = -jr- = 120. 

Example. — ^Find the value of x in — ^- + ^ = 4 -^ 

Solution.— The product of the denominators is 2 X 3 X 4 = 24. Mul- 

. • 1 • u ,, , , „, 24(x + 3) , 24a; _ , 24(5 - x) . 
tiplymg both members by 24, ^ ■ -\ — ^ = 96 H r , smce 

T — = H 2; — Since all the numerators are now divisible by the 

denominators, perform the divisions, thus clearing the equation of fractions, 
and obtain 12{x + 3) + 8a; = 96 + 6(5 - x), or 12* + 36 + 8x = 96 

90 
+ 30 — 6a;. Transposing and collecting terms, 26x = 90, or a; = ^ = 

3x®^. Ans. 



§2 MATHEMATICAL FORMULAS 21 

26. The subject of transformation of equations is of great 
importance in connection with formulas. For instance, consider 
the formula, 

.37wT 

p = 

V 

Suppose that it is desired to find another formula giving the 
value of w. Multiplying both members of this equation by v, pv 
= .S7wT, which may be written .37 Tw = pv. Dividing both 
members by .377^, 

pv 

^ = :37r' 

a formula that can be used to obtain the value of w. 

jp 
As another example, transform the formula t = .53L^/-^ so it 

can be used to find the value of P. Dividing both members by 

t IP 

.53L, the equation becomes —^ry = ^/-^; squaring both members 

.o6L \ o 

t^ P P 3.56^2 1 o t:« AT 

> smce nr.nn = o.oo — . JNow 



.532L2 S .2809L2 L^ ''"^"^ .2809 

multiplying both members by S, 

L^ ' 
A numerical factor may be transferred from the denominator 
to the numerator by multiplying the numerator by the reciprocal 

171 

of the factor. Thus, let a be the factor and — the fraction; then 
' an 

- X m - Xm ^ 

— = z = Since -^^^t^ = 3.56 very nearly, 

aw 1 , _ n .2809 

- XaXn 

a 

t' _ 3.56^^ t' ^ _±_ y i! ^ o rfi V ^ = 

.2809L2 ~ L2 ■ ^^' .2809L2 .2809 ^ L^ ^ L^ 

3.56^2 



27. Prime Marks and Subscripts. — In certain formulas, it is 
desirable to use the same letter for different measurements or 
quantities of the same kind. The formula given in Art. 1 is an 
instance. Here the outside diameter was represented by D and 
the inside diameter by d. Instead of using a capital letter for 
one diameter and a lower case letter for the other, the outside 
diameter might have been represented by d' and the inside 



22 ELEMENTARY APPLIED MATHEMATICS §2 

diameter by d". Had there been several other diameters used in 
the formula, they might be indicated by d'", d"" or ^^ ^ d!^ , etc. 
These marks are called prime marks, and they are read in connec- 
tion with the letters as follows: d prime, d second, d third, d 
fourth, etc. In connection with printing, these marks are called 
superior characters. 

When exponents are used in connection with letters affected with 
prime marks, the result is somewhat awkward (thus, d'^, d"^, 
d'"^, etc.) and it is then customary to use what are called sub- 
scripts (in printing, called inferior characters) , which are small 
figures of the same size as those used for exponents and placed 
slightly below and to the right of the letter; for example, d\ 
(read d sub-one), d^ (read d sub-two), dz (read d sub-three), etc. 
Exponents can be used with such letters very readily; thus, d\, 
d\, dl, etc. Such expressions are read d sub-one square, d sub-two 
cube, d sub-three fifth, etc. Always remember that letters having 
different subscripts or prime marks are just as different and are 
operated on in the same way as though different letters had been 
used instead. For instance, bd' + M" - 2d' = 3d' + 8d". 

28. Application of Formulas. — To apply a formula, substitute 
in the right-hand member the values of the letters and then per- 
form the indicated operations. 

Example 1. — Given the formula / = Ir^ j- + stt ) jr t- find the 



= (r^ - 



value of I when r = 12.3 and h = 4.8. 

Solution. — Substituting in the formula the values of r and h, I = (12.3* 

3 X 12.3 X 4.8 , 3 X 4.8^ \ 2 X 4.8 ^ ., oo aa oi, A.-i ^^a\ 
4 + —20") 3X12.3 -4.8 = (151.29-44.28+3.456) 

X .29907 = 110.466 X .29907 = 33.037, very nearly. Ans. 

^. , , •, o HA' + 2VA^" + 3A") ' ^ „ 

Example 2. — Given the formula o = , : — , nnd o 

4:{A'.+ VA'A" +A') 

when h = 14.5, A' = 40.84, and A" = 30.63. 

Solution. — Substituting in the formula the values given. 



„ ^ 14.5(40.84 + 2V40.84 X 30.63 + 3 X 30.63) 

4(40.84 + \/40.84 X 30.63 + 30.63) 

^ 1 4.5(40.84 + 70.737 + 91.89) ^ 14.5 X 203.467 ^ _ . 

4(40.84 + 35.3685 + 30.63) 4 X 106.8385 " ' 

PK 
Example 3. — Given the formula p = ^Mtr , t^ ' to find D when p = 56, 

^ DN + K f ' 

P = 200, K = 700, and N =150. 

Solution. — First transform the equation so D will stand alone in the 
left-hand member. Multiplying both sides of the equation by DN + K, 



§2 MATHEMATICAL FORMULAS 23 

-pND + pK = PK; transposing the term pK to the second member, pND 
— PK — pK; Dividing both members by pN , 

n PK-pK _ K{P - p) 

^ ~ pN pN ' 

Now substituting in the formula for D the values of K, P, p, and N, 

_ 700(200 - 56) _ 

^ - 56 X 150 - ^^- '^'''• 



+ f^ + i_" = 7 ^ _ |\ +35. Ans. x = llxV- 



EXAMPLES 

(1) Find the value of x in the equation 

13 - 5(3 + 4x) = 4x + 20 - 4(x + 20). Ans. x = 2.1 

(2) Find the value of x in the equation 

X 2x , 7x 
2"^ 3 

(3) Find the value of x in the equation 

X ^„ ,„ , a(78 - 45a) 

15a - 2ax + ^ = 26 - 17a:e. Ans. x = ^ — 

(4) When w' = 3, w" = 8.5, w'" = 3.6, s' = .0951, s" = 1, s'" = .1138 
f' = t" = 60, and t'" = 840, find the value of t in the formula. 

^ v^ + w"s^r^+w-s-t"' ^^^ ^ ^ 9,_,,3_^ 

w's' + w"s" + w"'s"' 

(5) When H = 178, Ta = 540, and Tc = 1210, find p from the formula 

p=ff(^^-^) ^^«- P = 1.343. 

(6) When s = }i{a + b + c), and a = 24.5, h = 37.8, c = 43.3, find r 
from the formula 

r = J (s -a)is -b){s -c) _ ^^^ ^ ^ 8.7394+. 

(7) Given the formula c = V2rh - h^, find c when r = 32| and h = llf . 

Ans. c = 24.955 -. 

(8) Given n = —, , find n when L = 1800, tt = 3.1416, d = 8, 

Vird^ + t^ 
3 
and ( = o2' ^^*- " ^ 126.94 -. 



QUADRATIC EQUATIONS 

29. To solve an equation is to find the value of the quantity 
represented by some particular letter; this quantity is called the 
unknown quantity. In the examples given in Art. 25, the un- 
known quantity is x. 

What is termed the degree of an equation is determined by the 
highest exponent of the unknown quantity. In the equations so 
far given, the highest exponent of the unknown quantity was 1 ; 



24 ELEMENTARY APPLIED MATHEMATICS §2 

hence, these equations are said to be of the first degree. An 
equation of the first degree is also called a linear equation. 

When the highest exponent of the unknown quantity is 2, as in 
the equation 3x + Qx^ — 12 = 7(2 -{- x^), which reduces to 2x^ 
+ 3x = 26 after the various transformations have been made, 
the equation is said to be of the second degree. An equation of 
the second degree is generally called a quadratic equation. 

The majority of equations that occur in practice reduce by 
transformation to linear or quadratic equations. Any linear 
equation may be represented by 

ax = b, (1) 

in which a and h may have any numerical value, and may be 
positive or negative. If a be negative, it may be rendered posi- 
tive by multiplying both sides of the equation by — 1 or, what is 
the same thing, dividing both sides (i.e., both members) by — 1. 
Thus, —ax = h divided by — 1 becomes ax = —b. 

Any quadratic equation will reduce to the form 
ax^ -\- bx = c, (2) 

in which a, b, and c may have any numerical value, and may be 
positive or negative. If a be negative, it may be rendered posi- 
tive by dividing both sides of the equation by — 1. Thus, —ax^ 
-\- bx = —c divided by — 1 becomes ax^ — bx = c. 

In equations (1) and (2), x is the unknown quantity whose 
value is to be found, and which may be represented by any letter 
whatever instead of x. For instance, the equation might be n^+ 
8n = 33; here n is the unknown quantity, a = 1, 6 = 8, and c = 
33. In the equation 3s^ — 13s = —14, the unknown quantity 
is s, a = 3, 6 = — 13, and c = — 14. 

30. It is sometimes convenient to use what is called the double 
sign which is written + or + . The first character is read plus 
or minus, and the second character is read minus or plus, the 
name of the upper sign being pronounced first, both characters 
being a combination of the plus and minus signs. Either charac- 
ter really represents two signs and is treated as two signs, whence 
the name, double sign. Thus, 16 ± 5 = 16 + 5 = 21 or 16 — 5 
= 11. 

Since +a X +a = + a^ and —aX—a= -}-a^, it follows 
that the square root of a^ may be either +a or —a; this fact is 
indicated by writing -y/a^ = ±a, consequently, 7 + -\/l6 = 11 
and 7 — \/lQ = 3 may be represented by 7 + \/l6 = 11 or 3. 



§2 MATHEMATICAL FORMULAS 25 

If the number whose square root is to be extracted is negative, 
the operation can only be indicated, since no negative number can 
be squared that will give a positive product. For this reason, 
such expressions as a/— 16, a/— c, etc. are called imaginary 
quantities. Since the square root of a negative quantity is in- 
dicated by V— «j it follows that the square of an imaginary 
quantity is a negative quantity; thus, (■%/— a)^ = —a, (\/— 16)^ 
= — 16, etc. 

31. Roots of Equations. — Any value of the unknown quantity 
that satisfies the equation is called a root of the equation. An 
equation of the first degree, a linear equation, has but one root, 
while an equation of the second degree, a quadratic equation, 
has two roots. 

To solve a linear equation, reduce it to the form ax = h; then, 

dividing both members by a, a; = -j and the result thus obtained 

is the root of the equation. 

To solve a quadratic equation, reduce it to the form 

ax^ -\- hx = c; 

then substitute the values of a, h, and c, with their proper signs, 
in the formula 

^ ^ -b ± \/b^ + 4ac 
2a 

This formula gives two values for the unknown quantity; that is, 
the equation has two roots. 

To apply the formula, let the equation be Ss^ — 13s = — 14. 
Substituting 3 for a, — 13 for h, and — 14 for c, 

_ -(-13) ±V(-13)2+4 X3 X- 14 

s — 



2X3 

13±Vl69 - 168 13 ± 1 



= 24 or 2. 



6 6 

If either of these values be substituted for s in the original 
equation, it will satisfy the equation. Thus, since 2| = y, 
3(1)2 - 13 X i = -^/ - -V- = - ¥- = - 14, and 3 X 2^ - 13 
X2=12 — 26=— 14. Since the first member equals the second 
member in both cases, both of the values obtained for s are said 
to satisfy the equation, and the roots of the equation are 2f and 2. 

The above formula giving the value of x is so important that it 
should be thoroughly committed to memory. 



26 ELEMENTARY APPLIED MATHEMATICS §2 

7 
Example 1. — Find the roots of the equation 2x —-z = 5(21 — x). 

Solution. — Removing the parenthesis and transposing to the first mem- 
ber the term containing x, 

7x ^ = 105 

X + 5 

Clearing of fractions, 7x^ + 35x — 7 = 105x + 525 
Transposing and collecting terms, 

7a;2 — 70x = 532 

Dividing by 7, x^ - lOx = 76 

-(-10) + V102 +4X1 X 76 



Substituting in the formula. 



2 X 1 
10 ± \/404 10 + 20.1 



2 2 

= 15.05 or -5.05. Ans. 

7 
Substituting 15.05 for x in the original equation, 2 X 15.05 — ^ - „_ . _ 

= 5(21 — 15.05). Performing the operations indicated, 29.75087+ = 

29.75, which agrees close enough for practical purposes. Substituting —5.05 

7 
forx, 2 X —5.05 — _r nc i c = ^[21 — (—5.05)]. Performing the opera- 
tions indicated, 129.9 = 130.25, which also agrees close enough for most 
practical purposes. In the first case, the difference between the two values 
was 00.00087, and in the second case, 000.35. If more accurate results 
are desired, the square root of 404 must be found to a greater number of 
deicmal places. In almost every case, only the positive root is desired, 
and the value found is accurate enough. 

X 4x^ 

Example 2. — Find the value of x in 5 — 4 — x^ + 2x ^ = 45 — 

o o 

3x2 4. ^^ 

SoLXJTioN. — Transposing the 4 to the right-hand member and the two 
terms containing x to the left-hand member, 

I + 2x2 - 2x - ^ = 49 (1) 

Clearing of fractions, 

5x + 30x2 _ 30a; _ 12x2 = 735 

Collecting terms, 18x2 - 25x = 735 

« u +•++••+;, . 1 25 ± \/252 + 4 X 18 X 735 

Substitutmg m the formula, x = 

^ X lo 

_ 25 + 231.398- 

36 

= 7.1221+ or -5.7333-. Ans. 

Substituting these values in equation (1), which is practically the same 

as the original equation, Zi^|^ + 2 X 7.12212 - 2 X 7.1221 - ^ ^ 7^12212 

= 49; performing the indicated operations, 48.999 = 49; difference = 00.001. 

Also, -^^333 ^ 2( -5.7333)2 _ 2 x -5.7333 - 4(-5-J333)2 ^ ^^, ^^^^ 

forming the indicated operations, 49.0004 = 49; difference = 00.0004. 
Both results are close enough for practical purposes. 



§2 MATHEMATICAL FORMULAS 27 

It is always a good plan to substitute at least one of the roots 
for the unknown quantity in the original equation in order to 
make sure that no mistake has been made in the work. 



EXAMPLES 

Find the value of x in the following equations: 

(1) x^ - 16a; = -63. Ans. x = 9 or 7. 

(2) 3x2 -{- 8x = 56. Ans. x = 3.1882+ or -5.8549-. 
(.3) 7x2 ^ 20x = 32. Ans. X = U or -4. 

(4) x2 - §x = li Ans. X = 1| or -|. 

(5) x3 + (19 - x)3 = 1843. Ans. x = 11 or 8. 

(6) 24.3x2 - 65.28X = 427.486. Ans. x = 5.74732 or -3.06091. 



ACCURACY m CALCULATION 

32. Significant Figures. — The significant figures of a number 
are those figures which begin with the first digit and end with the 
last digitJ For example, in the numbers 2304600 and .00023046, 
the significant figures are 23046. No attention is paid to the 
decimal point in connection with significant figures, and ciphers 
to the left of the first digit and to the right of the last digit are 
not considered. The significant part of a number is that part 
which is composed of its significant figures. In the numbers 
.005236, 52.36, and 5236000, the significant part is 5236, and 
these numbers all contain four significant figures. 

In practice, before any formula can be used, it is usually neces- 
sary to make one or more measurements. Thus, before the 
formula in Art. 1 can be used, it is necessary to measure the length 
and the inside and outside diameters of the hollow cylinder. If, 
however, it is assumed that the inside diameter is, say, 8 in., that 
the weight is, say, 490 lb., and it is desired to find the outside 
diameter, it may be readily calculated as follows : 

W = .7854m;Z (D + d) {D - d) = .7854 wl (D^ - d^), whence, 

^ = ^S^l+^- ^/ ^xT604X30 + «' = "•««^^: 
in. In practice, this would probably be taken as 12 in., but in 
any case, a measurement would have to be made before this 
result could be applied. 

What is true in the case just cited is also true in practically 
every case that occurs in the application of formulas; measure- 
ments of some kind — weight, length, area, or volume — must be 
made either before or after the formula is used. 



28 ELEMENTARY APPLIED MATHEMATICS §2 

33. Accuracy in Measurements. — In commercial transactions 
and in ordinary calculations pertaining to engineering matters, 
measurements are seldom accurate to more than three significant 
figures. A measurement is said to be correct to n significant figures 
when, if expressed to n figures, the figures following the nth. figure 
being considered as a decimal, the difference between the two 
numbers is greater than —.5 and less than +.5. Thus, the 
number 3.141593 expressed to four significant figures is 3.142; 
expressed to three or five significant figures it is 3.14 or 3.1416. 
Regarding these several numbers as integers and subtracting the 
original number from them, 3142 — 3141.593 = +.407, which is 
less than +.5; 314 — 314.1593 = —.1593, which is greater than 
-.5; and 31416 - 31415.93 = +.07, which is less than +.5. 
Hence, 3.141593 correct to 3, 4, or 5 significant figures is 3.14, 
3.142, or 3.1416. 

When a number is expressed to a certain number of significant 
figures, say n, and it is necessary to employ ciphers to make up the 
required number of n figures, then these ciphers are counted as 
significant figures. For instance, 4599.996 and 4600.003 when 
expressed to 6 significant figures are written 4600.00. 

For any measurement to be correct to n significant figm-es, it 
is necessary to know that the figures to the right of the nth figure 
form a number that is greater than —.5 and less than +.5. 
Suppose that 4 tons of coal (8000 lb.) are bought. The purchaser 
will not object if there is a pound less and the dealer will not 
object if there is a pound more; in fact, a difference of 5 or possibly 
10 pounds would not be noticed. To obtain 4 significant figures 
correct in this transaction would require that the amount of coal 
delivered be not less than 7999.5 lb. and not more than 8000.5 lb. 
But the consideration of half a pound of coal would be ridiculous 
in connection with a weight of this amount. To be correct to 
3 significant figures, the coal delivered must be not less than 7995 
lb. nor more than 8005 lb., and if the amount actually delivered 
came within these limits, the transaction would be considered 
quite accurate. 

What is true of a heavy weight is also true of a light weight. 
Chemists have balances so accurate that when the proper pre- 
cautions are taken, a thousandth of a grain may be weighed; but 
these balances are suitable for weighing only very small amounts, 
and a weight of even a tenth of a pound might injure them; 
possibly a hundredth of a pound would be the greatest amount it 



§2 



MATHEMATICAL FORMULAS 



29 



would be safe to use. Since a pound contains 7000 grains, a 
hundredth of a pound contains 70 grains, or 70,000 thousandths 
of a grain, that is, 5 significant figures. To make certain that the 
fifth figure was correct, it would be necessary that the balances 
detect a variation of half a thousandth of a grain, which would be 
beyond the limit of any except very special instruments ; therefore, 
the fifth figure could not ordinarily be determined, and the weigh- 
ing would be correct to only four significant figures. 

The same considerations govern any other kind of measure- 
ment, and the following law may therefore be laid down: 
measurements may he considered sufficiently accurate for practical 
purposes if correct to 3 significant figures; very accurate, if correct to 
4 significant figures; and extremely accurate, necessitating special 
methods and instruments, if correct to 5 or more significant figures, 

34. Accuracy in Numerical Operations. — Suppose it is desired 
to multiply 4343.944819 by 3.141593 and obtain the result correct 
to 5 significant figures. Using all the figures of both numbers, 
the work is shown at (a), and the product to 5 significant figures 



{a) 


(6) 




(c) 


4342.944819 


4342.94 + 


4'3'4'2.'9'4 + 


3.141593 


3.14159 
13028.82 


3.14159 


13028.834457 


13028.82 


434 2944819 


434 29 


4 


434 29 


173 71779276 


173 71 


76 


173 72 


4 342944819 


4 34 


294 


4 34 


2 1714724095 


2 17 


1470 


2 17 


39086503371 


39 


08646 


39 


13028834457 


13643.73 


68746 


13643.73 



13643.765042756667 | 

is 13644—. Now limiting both multiplicand and multiplier to 
one more significant figure than is required in the product, in this 
case, 5 + 1 = 6 figures, the work is shown at (6) , and the product 
to 5 significant figures is 13644 as before. Inspecting the work at 
(6) , it will be observed that if a line be drawn as shown cutting off 
all figures to the right of the first partial product, the result is 
13643.73, which becomes 13644 when reduced to 5 significant 
figures. Observe that in the multiplication, the left-hand digits 
of the multiplicand and multiplier are placed under each other and 
the first partial product is found by multiplying the multiplicand 
by the first figure of the multiplier. The second partial product 
is then written one place to the right of the first partial product; the 



30 ELEMENTARY APPLIED MATHEMATICS §2 

third partial product is written one place to the right of the 
second partial product; etc. This brings the partial products in 
the same relative positions insofar as the order of the figures of 
the product is concerned as would be the case if the multiplying 
were begun with the right-hand figure of the multiple. 

The work at (c) is very much the same as at (6) , except that as 
soon as the first partial product has been obtained, the right- 
hand figure of the multiplicand is cut off; it is considered, however, 
in multiplying, in order to ascertain how much to carry. In 
finding the second partial product, there is nothing to carry in 
this case because 1 X 4+ = 4+. Having found the second 
partial product, cut off another figure in the multiplicand; then, 
since 4 X 9 = 36, call it 40 and carry 4 to the next product, 
obtaining 4X2 + 4 = 12; write 2 and carry 1, Then, 4X4 
+ 1 = 17; write 7 and carry 1. 4 X 3 + 1 = 13; write 3 and 
carry 1. 4 X 4 + 1 = 17, which write. Now cut off another 
figure, and 1 X 434 = 434, which write. Cut off another figure, 
and since 5 X 4 = 20, carry 2. Then, 5 X 3 + 2 = 17; write 
7 and carry 1. 5X4+1 = 21, which write. Finally, cut off 
one more figure, and 9X3 = 27, which call 30 and carry 3. 
Then 9 X 4 + 3 = 39, which write. It will be observed that 
the first figure obtained in each partial product is written directly 
under the last figure of the first partial product. Adding the 
partial products as they stand, the sum is 13643.73; or 13644 
when reduced to 5 significant figures. 

Now counting the number of figures at (a), it will be found that 
109 figures were used; at (6), 64 figures were used; at (c), 44 figures 
were used. The result in all three cases is the same when the 
product is expressed to 5 significant figures. Not only is there a 
great saving in the number of figures employed, but there is very 
much less liability of making mistakes, and it is very much easier 
to check the work. In order, however, to perform the work as 
shown at (c), it is necessary to begin multiplying with the first 
(left-hand) figure of the multiplier. 

As another example, find the product of 230.2585093 and 
15.70796 to 5 significant figures. The work is shown in the 
margin. Instead of reducing the numbers to 5 + 1 = 6 signifi- 
cant figures, they are written as they stand and the multiplicand 
is limited to 6 significant figures by cutting off the remaining 
figures. The first partial product is 2302.59, because the seventh 
figure is 5+. It is always best to locate the decimal point in 



2302.59 


1151 


29 


161 


18 


1 


61 




21 




1 



§2 MATHEMATICAL FORMULAS 31 

the first partial product. To locate it in this case, note that if the 
2'3'0 '2'5'8'5093 multiplier were 1.57+ , the first partial 
15 70796 product would be 230.259; but the multi- 

plier is 10 times 1.57+ , and the first 
partial product is therefore 10 X 230.259 
= 2302.59. After multiplying by 9 and 
obtaining 21 for the fifth product, cut off 
one more figure of the multiplicand, and 
og,g go since 6 X 2 = 12, write 1 for the sixth 

partial product. This method of multi- 
plying is slightly more accurate than that at (c) in the preceding 
example, when both factors were reduced to 5 + 1 = 6 figures. 
The product correct to 5 significant figures is 3616.9. Ans. 
35. Division may be performed in a somewhat similar manner. 

Thus, to find the quotient of 
3616.89 -h 230.2585093 to 5 
significant figures, arrange the 
work as shown in the margin. 
Limit the divisor to 5 + 1 = 6 
figures by cutting off all figures 
beyond the sixth, and then cut off 
another figure in the divisor every 
time another remainder is found. 
There will evidently be two 
integral places in the quotient, 
since 3616.890 -=- 230.258+ = 
15.7+ . The quotient correct to 5 significant figures is 15.708. 
As another example, find the quotient of 13643.765 -^ 3.141593 

to 5 significant figures. Limiting 
13643. 7'6'5(3.'1'4'1'5'9'3 ,, .°. j j- -j j x 5 

^ the divisor and dividend to 6 

12566 37 4342.94+ ^ u xt, ■,• -j j -n 

figures each, the dividend will not 

contain the divisor; hence, instead 
of cutting off another figure from 
the divisor, add another figure to 
the dividend, and then proceed 
with the division as shown. The 
quotient evidently contains 4 in- 
tegral places, since the quotient of 
13643 -T- 3 will contain 4 integral 
places. The quotient correct to 5 
significant figures is 4342.9. Ans. 



3616 


,890( 
59 


'2'3'0. 


'2'5'8'5093 


2302 


15, 


.70795- 


1314 


30 








1151 


29 








163 


01 








161 


18 








1 


83 








1^ 


61 

22 

21 

1 

1 









1077 39 


942 48 


134 91 


125 66 


9 25 


6 28 


2 97 


2 83 


14 


12 


2 



32 ELEMENTARY APPLIED MATHEMATICS §2 

36. Constants and Variables. — A large majority of the formulas 
used in practice contain one or more quantities that remain the 
same no matter what the conditions are that govern the problem. 
For instance, the formula in Art. 1 contains the quantity .7854. 
Now, no matter what the values of D, d, I, and w are, the quantity 
.7854 remains the same; for this reason, it is called a constant. 
The other quantities in this formula are called variables, because 
their values vary or change for different cylinders. For all cast- 
iron cylinders, w = .2604, and, hence, for all cast-iron cyUnders, 
w is also a constant, its value being .2604. 

It has been shown that for multiplication and division, the 
numbers iised may be limited to one more figure than the number 
of significant figures required in the result without loss of accuracy. 
What is true in this respect of multiplication and division is also 
true of addition, subtraction, powers, and roots. Consequently, 
in all practical applications of formulas, the number of significant 
figures used in all the quantities contained in the formulas may be 
limited to one more than is contained in the constant having the 
smallest number of significant figures, and the result should be 
limited to the same number of significant figures as this constant 

contains. For example, the formula A = p ■ o r.^ contains 

two constants, both having 3 significant figures; hence, the values 
of G and P may be limited to 3 + 1 = 4 significant figures, and 
the result when found should be expressed to 3 significant figures. 
The formula d = L54aD + 2.6 contains two constants, one of 
which contains but two significant figures; hence, the values of a 
and D may be limited to 3 significant figures and the value of d 
when found should be limited to 2 significant figures. 

37. In the examples that follow, unless for some reason a 
special exception is made, aU applications of formulas will be 
made under the assumption that all the quantities, both constants 
and variables, have been limited to one more significant figure 
than the number of significant figures contained in the constant 
having the least number of significant figures; if there are no 
constants, then all quantities should ordinarily be limited to 5 
significant figures, and the results will be expressed to 4 significant 
figures. 

Example. — In the formula s = .„,^y > suppose that W = 1200 pounds, 

Z = 108 inches, E = 1,500,000 pounds, and / = j^, in which 6 = 3 inches 
and d = 8 inches; find the value of s. 



§2 MATHEMATICAL FORMULAS 33 

Solution. — In this formula, the constants 48 and 12 are exact, and are 
therefore correct to any nunaber of significant figures; the number 1,500,000 
is correct to only 2 significant figures, and it is doubtful even if the second 
figure is correct. Consequently, it is useless to employ more than 3 signifi- 
cant figures in the calculation, 

3X8^ 

First calculate the value of 7, obtaining I = — j^^ = 128. Substituting 

this value of / and the other values given in the above formula, 

9 
m P 27 27 

_ im X 1083 ^ i^^ X 19^ X m ^ 6561 ^ 

^ ~ ^^ X xmm X 128 mm x m 40000 ' ^ • • 

4 15000 20000 3^ Ans. 

2 
The formula just given gives the value of s in inches, s being 
the deflection of a beam having a certain shape and under certain 
conditions of loading, etc. Therefore, the deflection in this case 
would be taken as .16 inch, and this value is as close as can be 
obtained, although it might be a trifle more or a trifle less. Note 
that instead of cubing the number 108, the cube was expressed as 
the product of three factors, in order to employ cancelation. 



APPROXIMATE METHOD FOR FINDING ROOTS 

38. Cube and Fifth Roots of Numbers. — In certain formulas 
used in engineering, it is sometimes necessary to extract the cube 
root or, less frequently, the fifth root of a number. These roots 
may be found by an extension of the method explained in Arith- 
metic for finding the square root; but it entails a great amount of 
labor, and for practical purposes an approximate method answers 
the purpose equally well and is much easier to apply. Of the 
many approximate methods that have been recommended, the 
following is, perhaps, the simplest and most accurate. It was 
discovered by Charles Hutton, a famous English mathematician. 

Let n = the number whose root is to be found; let r = the 
index of the root, and let a be a number a little greater or a little 
less than the exact value of the root, so that a'' is a little greater 
or a little less than n. Then, 

r/- r(r + l)n + (r-lKl i /^ 

V w = 7 :r\ rn — , ^\ A ^^ nearly. (1) 

L(r — l)n + (r + ijoTA 

For cube root, r = 3, and formula (1) reduces to 

3/- /2n + a\ .^. 



34 ELEMENTARY APPLIED MATHEMATICS §2 

For fifth root, r = 5, and formula (1) reduces to 
5/- /3w + 2a\ 

The more nearly a*" approaches in value to n the more accurate 
will be the value obtained for the root. In order to save labor in 
finding a, the following table has been calculated; it gives the 
cubes and fifth powers of 1.1, 1.2, 1.3, etc. up to 9.9, andisusedas 
described below. 

Suppose it is desired to find the cube root of 34,586. The first 
step is to point off the number into periods of 3 figures each, 3 
being the index, beginning with the decimal point and going to 
the right and left. If the number contains an integral part, 
point off that part only; but if it is a pure decimal, point it off to 
the right. Thus, the above number, when pointed off, becomes 
34'586. If it had been .034586, it would become .034'586 when 
pointed off. 

The next step is to move the decimal point so that it will follow 
the first period that contains a digit, and the given number then 
becomes 34.586. The given number, after shifting the decimal 
point will be called the altered number. Of course, if the integral 
part of the given number contains not more than 3 figures, it is 
not necessary to shift the decimal point. 

Now referring to the table, and looking in the column headed 
n^, the number 34.586 is found to He between 32.768 = 3.23 ^nd 
35.937 = 3.33; the cube root of 34.586 therefore lies between 3.2 
and 3.3, and one of these two numbers is to be selected for a in 
applying formula (2) . To decide which one, find which of the 
two cubes just mentioned is nearest in value to the altered 
number. The easiest way to do this is to 
add the two numbers and divide the sum 
by 2 (this is the arithmetical mean of the 
two numbers). If the altered number is 
less than the arithmetical mean, use the 
smaller number, but if it is larger, or if it 
is near the arithmetical mean in value, use 
the larger number. In the present case, 
(32.768 + 35.937) -^ 2 = 34.3525; hence, 
use the larger number. Then a = 3.3 and 
a^ = 35.937. The work of applying the 
formula is shown in the margin. After 
performing the division, the quotient is 



34.586 
2 


35.937 

2 


69.172 
35.937 


71.874 
34.586 


105.109 ( 


106.460 


95 814 


.98731 


9 2950 


3.3 


8 5168 


2.96193 


7782 


296193 


7452 


3.258123 


330 

319 

11 





§2 



MATHEMATICAL FORMULAS 
CUBES AND FIFTH POWERS 



35 



n 


n3 


n6 


n 


n3 


ns 


1.0 


1.000 


1.00000 


5.5 


166.375 


5032.84375 


1.1 


1.331 


1.61051 


5.6 


175.616 


5507.31776 


1,2 


1.728 


2.48832 


5.7 


185.193 


6016.92057 


1.3 


2.197 


3.71293 


5.8 


195.112 


6563.56768 


1.4 


2.744 


5.37824 


5.9 


205.379 


7149.24299 


1.5 


3.375 


7.59375 


6.0 


216.000 


7776.00000 


1.6 


4.096 


10.48576 


6.1 


226.981 


8445.96301 


1.7 


4.913 


14.19857 


6.2 


238.328 


9161.32832 


1.8 


5.832 


18.89568 


6.3 


250.047 


9924.35643 


1.9 


6.859 


24 . 76099 


6.4 


262.144 


10737.41824 


2.0 


8.000 


32.00000 


6.5 


274.625 


11602.90625 


2.1 


9.261 


40.84101 


6.6 


287.496 


12523.32576 


2.2 


10.648 


51.53632 


6.7 


300.763 


13501.25107 


2.3 


12.167 


64.36343 


6.8 


314.432 


14539.33568 


2.4 


13.824 


79 . 62624 


6.9 


328.509 


15640.31349 


2.5 


15.625 


97.65625 


7.0 


343.000 


16807.00000 


2.6 


17.576 


118.81376 


7.1 


367.911 


18042.29351 


2.7 


19.683 


143.48907 


7.2 


373.248 


19349.17632 


2.8 


21.952 


172.10368 


7.3 


389.017 


20730.71593 


2.9 


24.389 


205.11149 


7.4 


405.224 


22190.06624 


3.0 


27.000 


243.00000 


7.5 


421.875 


23730.46875 


3.1 


29.791 


286.29151 


7.6 


438.976 


25355.25376 


3.2 


32.768 


335.54432 


7.7 


456.633 


27067.84157 


3.3 


35.937 


391.35393 


7.8 


474.552 


28871 . 74368 


3.4 


39.304 


454.35424 


7.9 


493.039 


30770.56399 


3.5 


42.875 


525.21875 


8.0 


512.000 


32768.00000 


3.6 


46.656 


604.66176 


8.1 


531.441 


34867.84401 


3.7 


50.653 


693.43957 


8.2 


551.368 


37073.98432 


3.8 


54.872 


792.35168 


8.3 


571 . 787 


39390.40643 


3.9 


59.319 


902.24199 


8.4 


592.704 


41821.19424 


4.0 


64.000 


1024.00000 


8.5 


614.125 


44370.53125 


4.1 


68.921 


1158.56201 


8.6 


636.056 


47042.70176 


4.2 


74.088 


1306.91232 


8.7 


658.503 


49842 . 09207 


4.3 


79.507 


1470.08443 


8.8 


681 . 472 


52773.19168 


4.4 


85.184 


1649.16224 


8.9 


704 . 969 


55840.59449 


4.5 


91.125 


1845.28125 


9.0 


729.000 


59049.00000 


4.6 


97.336 


2059.62976 


9.1 


753.571 


62403.21451 


4.7 


103.823 


2293.45007 


9.2 


778.688 


65908.15232 


4.8 


110.592 


2548.03968 


9.3 


804.357 


69568.83693 


4.9 


117.649 


2824.75249 


9.4 


830.584 


73390.40224 


5.0 


125.000 


3125.00000 


9.5 


857.375 


77378.09375 


5.1 


132.651 


3450.25251 


9.6 


884.736 


81537.26976 


5.2 


140.608 


3802.04032 


9.7 


912.673 


85873.40257 


5.3 


148.877 


4181.95493 


9.8 


941.192 


90392.07968 


5.4 


157.464 


4591 . 65024 


9.9 


970.299 


95099 . 00499 


5.5 


166.375 


5032.84375 


10.0 


1000.000 


100000.00000 



36 ELEMENTARY APPLIED MATHEMATICS §2 

multiplied by a = 3.3, and the product is the cube root of 34.586, 
approximately. Since the decimal point was shifted one period to 
the left to form the altered number, it must be shifted one place 
to the right in the root, and v^ 34,586 = 32.58123, approximately. 
The root to 8 significant figures is 32.581178; therefore, the root 
as found by the formula was correct to six significant figures. 

38. The root when calculated as just described will always be 
correct to at least five significant figures, except when the altered 
number is less than 1.331 = l.P and is nearly equal to the arith- 
metical mean of 1 and 1.331. This case will be discussed in 
Art. 39. 

If for any reason, more figures are desired than can be obtained 
by proceeding as above, express the root as found to 3 or 4 signifi- 
cant figures, and substitute it for a in this formula. Thus, in the 
preceding example, the root to 4 figures is 32.58; substituting this 

f • f 1 ^o^ V^T^^ / 2 X 34586 + 32.58^ \ ,, ,, 

for a m formula (2), V34,586 = \ 34536 -|- 2 X 32 58V 

= 32.581177737944— ; the root correct to 14 significant figures is 
32.581177737942—; hence, the root as calculated was correct to 
13 significant figures. 

The reason for shifting the decimal point to get the altered 
number is obvious — to make it easier to use the table. Since 
■^34586 = ^1000 X 34.586 = 10^34.586, the decimal point 
must be moved one place to the right after the root of the altered 

number has been found. Similarly, -v^. 034586 = \/"17)nn" 

= TTrV^34.586; consequently, if the decimal point is shifted one 
period to the right to form the altered number, it must be shifted 
one place to the left in the root. 

39. When the altered number is less than 1.2, the fifth figure of 
the root as calculated by the foregoing method will usually be 
incorrect. In such cases, proceed as follows: Suppose the cube 
root of .001166 be desired. Pointing off, the result is .001 '166. 
Shifting the decimal point one period to the right, the altered 
number is 1.166, which is less than 1.2. Now divide the decimal 
part by the index of the root, and the quotient is .166 -^ 3 = 
.055+, which is nearly equal to the decimal part of the cube root 
of 1.166, the integral part being 1; that is, v^l.166 = 1.055, 
nearly. Substituting this value for a in formula (2), 

v'Hee = P ^.}''!^l t ! n^S ) 1-055 = .997655025 x 1.055 = 

V1.I66 + 2 X 1.0553/ 1.052526051. . 



§2 



MATHEMATICAL FORMULAS 



37 



The root correct to 12 significant figures is 1.05252604197. 
Hence, ^.001166 = .1052526+, by the formula. 

40. The fifth root of a number is found in exactly the same 
manner, using formula (3). The only difference in the process is 
that the number must be pointed off into periods of five figures 
each, because the index of the root is 5. As an example, find the 
fifth root of 214.83. Here the integral part of the number 
contains only 3 figures, and it is not necessary to point off the 
number. Referring to the table, the given number falls between 
2.9^ = 205.11149 and 3.0^ = 243; the arithmetical mean of 
these two numbers is 224.055745, and as this is greater than 
the given number, use 2.9 for a in formula (3). Substituting 
in the formula, 



5/xrj^ ^ /3 X 214.83 + 2 X 205.11149\ 
'^^^'^•^^ \2 X 214.83 + 3 X 20531149/ 



214.83 214.83 
3 2 



644.49 429.66 
205.11149 205.11149 
2 3 



410.22298 

644.49 
1054.71298 
1044 99447 


615.33447 
429.66 
(1044.99447 
1.00930006 


9 71851 


2.9 


9 40495 
31356 


2.01860012 
908370054 



31350 



2.926970174 



2.9 = 2.92697+. 

Ans. 

The whole calculation is 
shown in the margin. It 
was really not necessary to 
use BO many decimal places, 
but since but very little 
labor would be saved by 
using a smaller number, it 
was not considered worth 
while to abbreviate the 
process further. In general, 
however, only one more signifi- 
cant figure would be used than 
was desired in the root. 



6 



As in the case of cube root, the method will give at least five 
significant figures correct, except when the altered number lies 
between 1^ = 1 and 1.1^ = 1.61051 and is near the arithmetical 
mean of these two numbers. In such cases, proceed as described 
in connection with cube root, dividing the decimal part of the 
altered number by the index, in this case 5, to find the decimal 
part of the root. For example, find the fifth root of 1.308375. 
Here .308375 -^ 5 = .061675, and -y^l. 308375 = 1.06 nearly. 
Substituting 1.06 for a in formula (3), 

.VT^nsW^- / 3Xl.308 3 75+2Xl.06n 
Vl.dUbd/i) \,2X 1.308375+3X1.06=;' 



1 .06 = 1.055197 -, say 1.0552. Ans. 



38 ELEMENTARY APPLIED MATHEMATICS §2 

The root correct to 9 significant figures is 1.05522833. If 
more figures of the root are desired, substitute 1.055 for a. 

41. The fourth root is very seldom required. In case it should 
be necessary to find the fourth root, all that need be done is to 
extract the square root and then exti'act the square root of the 
result; in other words, -x/n = S-y/n-. For instance, v^97.34 = 
N/ a/97.34 = ^9.8661 = 3.1410 +• Here the square root of 
97.34 = 9.86610+ , and the square root of 9.8661 = 3.1410+ . 

42. If the table of cubes and fifth powers is not available, find 
the first figure of the root by trial; then substitute this value for 
a in formula (2) or (3) and calculate the root to three figures, 
calling the result h. Express b to two figures, and substitute it 
for a in formula (2) or (3) ; the remainder of the work is the same 
as before, the value of a being calculated instead of being taken 
from the table. 

Referring to the example of Art 37, the altered number is 
34.586. To find the first figure of the root, note that 3^ = 27 

27 -j- 64 
and 4^ = 64; the mean of these two powers is ^ = 45.5; 

hence, use 3 for the first figure of the root. Then, a = 3 and 
a3 = 33 = 27, and 

-.y^T^eft - / 2 X 34.586 + 27 \ „ _. 

^^^•^^^ - ( 34.586 + 2 X 27 ) = ^-2^- = ^- 
Expressing this root to two figures, b = 3.3. Substituting this 
value of 6 for a. in the formula, 3.3^ = 35.937, and 

It will be observed that insofar as the first five significant 
figures are concerned, either 3.2 or 3.3 may be substituted for 
a in the formula; but 3.3 is a slightly better value in this case, since 
it gives the root correct to 6 figures, while 3.2 gives only 5 
figures correct. 

The same procedure will be followed in the case of fifth roots; 
thus, referring to Art. 40, the altered number is 214.83. Here 
3^ = 243 and 2^ = 32; obviously, 3 is the proper number to sub- 
stitute for a in formula (3), and 

Expressing b to two figures, the root is 2.9, which is the same as 
the value of a used in Art. 40. 



§2 MATHEMATICAL FORMULAS 39 

This method may be appHed to the special cases of Arts. 39 
and 40. Thus, to find the value of -v^l.166, the first figure of the 

root is obviously 1, and L .^^'1 o v 1 ) ^ ^ 1-0524. Now 

when the altered number does not differ greatly from a^, the 
value of b may be calculated to 4 or 5 significant figures, and h 
may be expressed to 3 or even 4 figures, if desired, before sub- 
stituting for a in the formula; but, unless great accuracy is 
desired, it is best to express 6 to 3 figures, to save labor in calcu- 
lation. In the present case, substitute 1.05 for a in the formula. 
Similarly, in Art. 40, the altered number is 1.308375, and the 
first figure of the root is 1. Hence, 

5/T^7^^5^ / 3 X 1.308375 + 2 X 1 \ . , „_.„ 
^1^08^75 = ( 2 X 1.308375 + 3 X 1 ) ^ = ^'^^^^^ 

In this case, 1.05 or 1.055 may be substituted for a. 
If desired, this method may be used for finding the square 
root of numbers instead of employing the exact method de- 
scribed in Arithmetic. For square root, r = 2, and formula 
(1) of Art. 36 becomes 

^3n + a^\ 
{n + SaV 
As an example, find the square root of 3.1416. Here a is 

evidently 2, and ( 3 1415 _l 3 v^ 4 ) X 2 = 1.773+ . Using the 

first three figures for a, V3.1416 = 3 ^^^g + 3 x 1 77^ ^ 

1.77245594 — . By the exact method, the root to 9 significant 
figures is 1.77245592; hence, the root as found by the formula 
was correct to 8 significant figures. 

The exact method is so easy to apply, that the reader is ad- 
vised to use it in preference to the formula. 



/- /3n + a\ 
Vn = (;^^r^) 



EXAMPLES 

Calculate the roots in the following examples to 5 significant figures: 

(1) v^.04608. Ans. .35851+. 

(2) ^.86402. Ans. .97119+. 

(3) v^l.0947. Ans. 1.0306 +. 

(4) -v/324, 096,815. Ans. 686.90-. 

(5) -v/4, 063,972. Ans. 20.979+. 

(6) v^3.1416. Ans. 1.4646 -. 



ELEMENTAEY APPLIED 
MATHEMATICS 

(PART 1) 



EXAMINATION QUESTIONS 

(1) Explain the difference between a coefficient and an ex- 
ponent. (&) What is meant by the reciprocal of a number or 

2 , c 

quantity? (c) What is the reciprocal of a — - ? Ans. _ ^ . 

(2) Add a — b,b — c, c — d, and d — c. Let a = 20,h = 16, 
c = 11, and d = 8 and prove the result obtained was correct. 

Ans. a — c. 

(3) Find the sum of: a^ - Zax^ - daz^ + xh"" + z^ dax^ 
- 4a3 + 2a;22 + Qaz^ - x\ 2,a^ + 2a^z^ - 2xh^ + Zah - 5az^, 
4:X^ - 6aH - Sa^ - Sz^ + xH\ 

Ans. Sx^ - ^aH + 2xz^ + Sa^g + 2aH^ - 2az^ - a^ - 2a^ - 2z\ 

(4) Subtract fa + 36 + io; + ^ax from |a - 76 - 3aa; + ix. 

Ans. a + ix — iax — 112. 

(5) Multiply a^ - lOa^b + 40a362 _ gOa^&s _|_ soa?)^ - 326 ^ 
by a — 26. 

^ns. a^ - 12a56 + 60d%'" - lQOa%^ + 2400^6* - 192a65 + 6466. 

(6) By performing the indicated operations, reduce the follow- 
ing expression to a simpler form : 

a2(3a - 5a;) - (2a - x^ia - 2x) + (a^ -\- ax + x^)ia - x) - x\ 

Ans. ax {7 a — 9x). 

(7) Divide x^ - Ax^ - Slrc^ - 6x + 120 by a; + 5. 

Ans. x^ - 9x^ - 6a; + 24. 

. , P • 3 - 11a; .,, , 

(8) Change the signs of the fraction ~ ^ i g^ _ q without 

3 - 11a; 
changing the value of the fraction. Ans. g _ g^ _ ^^2 

(9) Given the formula v ^ Vo - ^at^, rewrite it so it will ex- 
press (a) the value of a, and (6) the value of t. 

2{vo - v) 



a= ^, 



ins. 



l2{vo - v) 
41 



12 {vo - 
\ a 



42 ELEMENTARY APPLIED MATHEMATICS §2 

(10) Given the formula 

23 1 ,00155 

_ n s /- 

''".552l(23 + ^5l55V^''^'' 
\ s / r 

find the value of v when n = .013, s = .005, and r = .559, 

Ans. V = 3.6+. 

(11) Remove the signs of aggregation from 

(27 - 4:x)[40 - 5(3a: + 7)] + 6{24 - (x + 2)(x - 4)[3 

- 5x(4 + a;)]} 
Ans. dOx^ + 60a;3 - 438a;2 - 1349a; + 423. 

(11) Divide 40x' - 700;^ + Ax^ + 724^3 + 88a;2 _ loSOrc 
- 3390 by 5x^ - 20a; + 48. 

Ans. 8x' + 18a;3 - 4a;2 - 44a: - 120 - J'l^\~ ^f '^?q - 

5x^ — 20a; + 48 

(12) Find the value of y in the equation. 

o _L 4?/ y c /% \ 

^^ + T - 2 = My - ^ + ^j 

Ans. 7/ = 4.0223+ 

(13) Find the two values of x in the equation lla:^ — 35a; = 40 

Ans. x = 4.0743+ or -.89251- 

(14) What is the cube root of 705.33> Ans. 8.9015+ 

(15) What is the fifth root of .76054? Ans. .94673- 

(16) What is the value of \/26.252 + 17.52? Ans. 31.559+ 



ELEMENTARY APPLIED 
MATHEMATICS 

(PART II) 



MENSURATION OF PLANE FIGURES 



DEFINITIONS 

43. Mensuration deals with the measurement of the length of 
lines, the area of surfaces, and the volume of solids; its principles 
and rules are used in every occupation and industry. The 
subject is, therefore, of the greatest importance. 

Every material object occupies space in three directions — in 
length, in breadth or width, and in thickness or depth. Every 
magnitude or body is consequently said to have three dimensions 
— length, breadth, and thickness. 

44. A mathematical line indicates direction only; it has only 
one dimension, length, and has no breadth or thickness. Such 
a line could not be seen; hence, every visible Hne, no matter how 
fine it may be, has breadth, but when considered in connection 
with problems relating to mensuration, the breadth of all lines 
is disregarded and they are conceived as having only length. 

45. A straight Line is one that extends continuously in one 
unvarying direction. In mathematics, straight lines are assumed 

to be infinite in length; that is 

they are assumed to be capable "^ ~ ~ 

of being extended in either 

direction without limit. Fig. 3 shows a straight hne, and it is 
assumed that it can be extended to the right or to the left as far 
as is desired — one foot, a mile, a hundred thousand miUion miles 
or farther — without any change in its direction. In practice, 
only short parts, called segments, of straight lines are con- 
sidered; and to distinguish one line from another, letters or 
figures are placed at the ends of the segments. Thus, in Fig. 3, 

43 



44 



ELEMENTARY APPLIED MATHEMATICS 



§2 



the line there shown is called the line AB, when it is considered 
as extending from A to B, or the line BA, when it is considered 
as extending from B to A. In mathematics, a straight hne is 
called a right line, and wiU usually be so designated hereafter. 

Another definition of a right hne is : a right hne is the shortest 
path between two points. This definition is obvious, since if 
the path deviates, even in the sHghtest degree, from a point to 
another point, the path (hne) will be longer than if it extended 
straight from one point to the other. 

46. A broken line is one that is made up of two or more right 
line segments; see Fig. 4. A broken Hne is distinguished by 
placing a letter (or figure) at the ends of each segment. Thus, 
the broken line in Fig. 4 is called the line ahcdef. 

A 




Fig. 4. 

47. A curved line or curve is one that has no right hne seg- 
ment; its direction changes continually throughout its entire 
length. See Fig. 5. A curve may be finite (hmited) in length 
or infinite (unhmited) in length, but in either case, no part of it 
is ever straight. An example of a finite curve is a circle. Fig. 
5, and an example of an infinite curve is a parabola, see (a), 





Fig. 6. 

Fig. 6. When only a portion of a curve is considered, it is called 
an arc, the word arc meaning bow, alluding to its shape. 

Curves are distinguished by placing letters at the ends of the 
arc and at such other points between as may be deemed advisable. 
Thus, in Fig. 5, the circle may be designated as ABCD; in (a), 



§2 MENSURATION OF PLANE FIGURES 45 

Fig. 6, the curve may be called abc; in (6), Fig. 6, the curve may 
be referred to as 12345. 

48. When two lines cross or cut each other, they are said to 
intersect, and the place where they intersect is called the point 
ot intersection. Thus, in Fig. 4, h, c, d, and e are points of in- 
tersection of the hnes ah and ch, of he and dc, etc., assuming that 
these segments of right hnes are prolonged (produced); in Fig. 
5, the points A, B,C, and B are the points of intersection of the 
right hnes AC and DB with the circle. 

49. In mathematics, a point indicates position only; it has no 
dimensions. In practice, a point resembles a very small circle — • 
more properly, a square — bub is always considered to be without 
dimension. A point may always be conceived as being the point 
of intersection of two lines. 

50. Parallel lines are right lines that are always the same 

distance from each other; ^ 

they never intersect no mat- ^ 

^ j^ — __- jj 

ter to what length they may p^^ ^ 

be produced. Thus, in Fig. 

7, the right hnes AB and CD are parallel. 

51. An angle is the difference in direction between two right 
hnes that intersect. The point of intersection is called the 

vertex of the angle, and the lines are 
called the sides. Angles are designated 
by placing a letter at the vertex and by 
two other letters, one on each line. In 
Fig. 8, the right lines AC and DB, inter- 
secting in the point 0, form four angles, 
designated as AOB, BOC, COD, and 
DO A. Two angles on the same side of 
a hne (which forms one of their sides) 
and separated by a common side are called adjacent angles. 
In Fig. 8, AOB and AOD are adjacent angles; AOB and BOC 
are also adjacent angles; etc. Note that adjacent angles have 
a common vertex, a common side, and the other side of both 
angles lies in the same right line. 

52. When two right hnes intersect in such a manner that the 
adjacent angles are equal, the hnes are said to be perpendicular 
to each other. For example, in Fig. 9, CD has been so drawn 




46 



ELEMENTARY APPLIED MATHEMATICS 



§2 



FiQ. 9. 



that all the adjacent angles, COB and CO A, COB and BOD, etc. 

are equal; hence, AB is said to be perpendicular to CD, and CD 
fj is said to be perpendicular to AB. All 

four angles, AOC, COB, BOD, and DO A 
are equal, and each is called a right angle. 

_^ 53. A horizontal line is one that is 
parallel to the horizon or to the water 
level; with the book held in its usual 
position for reading, AB, Fig. 9, is a 
horizontal line. Any line that is perpen- 
dicular to a horizontal line is a vertical 

line. In Fig. 9, CD is a vertical line. All vertical Unes have 

the same direction as a plumb line. 

54. An angle that is smaller than a right angle is called an 
acute angle. In Fig. 10, the angle AOB is evidently smaller 
than the right angle COB; hence AOB is an acute angle. 

If an angle is greater than 
a right angle, it is called an 
obtuse angle. In Fig. 11, 
AOB is an obtuse angle, 
because it is greater than 
the right angle COB. In 
Fig. 8, AOB and COD are 
acute angles, and AOD and BOC are obtuse angles. 

65. Angles are measured in several ways, the most common 
method being to use an arc of a circle. This method is called the 
angular measure of angles, and the table of angular measure was 
given in Arithmetic. The entire circle is supposed to be divided 
into 360 equal parts called degrees, the abbreviation for which is 
(°) ; each degree is then divided into 60 equal parts called minutes, 
the abbreviation for which is (') ; and each minute is divided into 
60 equal parts called seconds, abbreviation (")• 

The draftsman measures an angle by means of an instrument, 
usually made of metal or paper, called a protractor, which is a 
half circle divided into degrees and half degrees or degrees and 
quarter degrees. Evidently, the length of the sides of an angle 
have nothing to do with the size of the angle, since lengthening 
or shortening the sides does not change the direction of the lines. 
Consequently, by laying the protractor on the angle in such a man- 
ner that the center of the half circle coincides with the vertex of 




-B 



Fig. 10. 



Fig. 11. 




§2 



MENSURATION OF PLANE FIGURES 



47 



the angle and the bottom, or straight, side of the protractor 
coincides with one side of the angle, the other side will cross the 
arc of the half circle and the size of the angle can be read on the 
graduated edge. This is clearly shown in Fig. 12. Here the pro- 
tractor P is so placed that its center coincides with the vertex 
of the angle, and the bottom side, or edge, of the protractor coin- 
cides with the side OB of the angle; the other side, OA, of the 




Fig. 12. 



angle crosses the arc at 59", which is the size of the angle. More 
accurate instruments are in use, but they all are based on the 
same principle. 

55. A surface has no thickness. When a surface is perfectly 
flat, so that a right line may be drawn on it in any direction and 
anywhere on it, the surface is called a plane. A plane surface 
may be tested by laying a straightedge anywhere on it; if the 
straightedge coincides with the surface throughout the length of 
the straightedge, no matter where it is placed, the surface is a 
plane surface. 

Planes, like right lines, are supposed to be infinite in extent; 
that is, they are infinite in length and infinite in breadth; but, 
like lines, only parts of planes are considered in practice. 

56. Two planes intersect in a right line. For example, referring 
to Fig. 13, the planes ABCD and EFGH intersect in the right Hne 
MN, which is called the line of intersection. 

If from any point o in the line of intersection MN a line oa 
be drawn in the plane EFGH, perpendicular to MN, and a line 
ob be drawn in the plane ABCD, also perpendicular to MN, 
the angle aob is the angle which the planes make with each other. 



48 



ELEMENTARY APPLIED MATHEMATICS 



§2 



If this angle is a right angle, in which case, oa and oh are perpen- 
dicular to each other, the two planes are then said to be perpen- 




Fig. 13. 



dicular to each other; otherwise, they make an acute or obtuse 
angle with each other, according as the angle aoh is acute or 
obtuse. 



PLANE FIGURES 

57. A plane figure is any outline that can be drawn on a plane 
surface; and, since a plane is infinite in length and breadth, a 
plane figure may be of any size and any shape. To be complete, 
however, the figure must close; that is, if the figure be traced 
by moving the pencil over it, then, starting from any point on 
the outline and passing over the entire figure, the pencil must 
return to the point from which it started. 

58. Polygons. — The simplest plane figures are those made up 
entirely of right lines — called polygons, which means many angles, 
from 'poly (meaning many) and gonia (meaning angles), and so 
called because a 'polygon always has as many angles as it has sides. 

The right fines that form the outline of a polygon are called the 
sides of the polygon, and the angles included between the sides 
are called the angles of the polygon. The sum of the lengths of 
the sides is called the perimeter of the polygon. The perimeter, 
therefore, is the length of the outline that bounds the polygon; 
it equals the distance around it. 



§2 



MENSURATION OF PLANE FIGURES 



49 



59. Polygons are named in accordance with the number of 
sides that they have. A polygon with three sides is called a 
triangle; one with four sides is called a quadrilaterial; one of five 
sides is a pentagon; one of six sides is a hexagon; one of seven sides 
is a heptagon; one of eight sides is an octagon; one of nine sides 
is a nonegon; one of ten sides is a decagon; one of eleven sides 
is an undecagon; one of twelve sides is a dodecagon; etc. In 
practical work, the triangle, quadrilateral, and hexagon are very 
freely used, and the pentagon, octagon, and decagon are oc- 
casionally used; the other polygons are practically never used. 

60. A regular polygon is one in which all the sides and all the 
angles are equal. Fig. 14 shows regular polygons of 3, 4, 5, 6, 8, 




B 



C A 



D 




Quadrilateral 



Pentagon 




Hexagon 



Octagon 
Fig. 14. 



Decagon 



and 10 sides. In each figure, the sides are all equal in length, and 
the angles (called the interior angles) are all equal. Thus, in 
the regular pentagon, AB = BC = CD = DE = EA, and ABC 
= BCD = CDE = DEA = EAB. 

61. To find the number of degrees in one of the equal angles 
of a regular polygon, let n = the number of sides and a° =the 
number of degrees in one of the equal angles; then 

^o ^ (.n- 2) ^ ^g^o 

n 
For instance, in the regular triangle, the interior angles are 

equal to ^ — X 180° = 60°; in the regular quadrilateral, 

4—2 6—2 

a° = — J — X 180° = 90°; in the regular hexagon, a° = — ^ — 

X 180° = 120°; etc. 

4 



50 



ELEMENTARY APPLIED MATHEMATICS 



§2 




TRIANGLES 

62. Triangles are classified in two ways: according to their 
angles, and according to their sides. "When the angles are con- 
sidered, triangles are called right triangles, when they have one 
right angle; oblique triangles, when they have no 
right angle; and equiangular triangles, when all the 
angles are equal. An equiangular triangle is also 
an oblique triangle. Fig. 15 shows a right triangle, 
the right angle being at B. It may here be re- 
marked that when there is no possibihty of misun- 
derstanding, angles may be named by a single 
letter placed at the vertex. Thus, in Fig. 15, 
angle C is the angle ACB, angle A is the angle CAB, and the 
angle B is the angle ABC. 

When the sides are considered, triangles are called isosceles, 
scalene, or equilateral triangles. An isosceles triangle is one that 
has two equal sides, see Fig. 16; here CA = CB. A scalene 
triangle is one in which all the sides have different lengths, see 
Fig. 17. An equilateral triangle is one in which all the sides are 
equal, see Fig. 18. A scalene triangle is always an oblique tri- 
angle. An equilateral triangle is also an isosceles triangle; it 



Fig. 15. 




B A 




B A 




Fig. 16. 



Fig. 17. 



Fig. 18. 



is likewise an equiangular triangle. A right triangle is usually 
a scalene triangle, but if the two sides that enclose or form the 
right angle are equal, it is then an isosceles triangle also. 

For convenience in mathematical operations, triangles are 
lettered with capital letters at the vertexes and with small letters 
(lower case letters) placed at the middle of the sides, the lower 
case letters being in each instance the same as the capital letters 
at the angles opposite the sides. Thus, referring to Figs. 15-18, 
side a is opposite angle A, side 6 is opposite angle B, and side c 
is opposite angle C. 



§2 MENSURATION OF PLANE FIGURES 51 

The side on which any triangle (or any polygon) is considered 
as standing, the plane of the figure being supposed to be vertical, 
is called the base. In Fig. 14, AC is the base of the triangle; 
AD is the base of the quadrilateral; and AE is the base of the 
pentagon. In Fig. 15, CB is the base; in Figs. 16-18, AB is the 
base. 

63. Some Properties of Triangles. — (1) In any triangle, the 
sum of the three angles is always 180°; thus, in Figs. 15-18, 
A -{- B -\- C = 180°. Therefore, if two of the angles are known, 
the third can be found by subtracting the sum of the two known 
angles from 180°. Thus, in Fig. 17, if A = 68° 23' and B =24° 
35', C = 180°— (68° 23' + 24° 35') = 87° 2'. 

(2) In a right triangle, one of the angles being a right angle 
and therefore equal to 90°, the sum of the other two angles must 
be 180° — 90° = 90°; consequently, both angles must be acute, 
since neither can exceed (nor even equal) 90°. Also, this sum 
must be 90°; because the sum of the three angles is 180°, and 
since one of the angles if 90°, the sum of the other two angles 
must be 180° — 90° = 90°. Hence, if one of the acute angles of 
a right triangle is known, the other can be found by subtracting 
the known angle from 90°. Thus, in Fig. 15, suppose that the 
angle A = 28° 42'; then the angle C = 90° —28° 42' = 61° 18'. 

(3) The longest side is always opposite the greatest angle, and 
the greatest angle is always opposite the longest side. In Figs. 
15 and 17, 6 and c are respectively the longest 
sides; hence, B and C are respectively the 
largest angles. 

(4) If any two sides of a triangle are equal, 
the angles opposite those sides are also equal. 
In Fig, 16, a and b are equal; hence A = B. 
Consequently, in every isosceles triangle, two 
of the angles are equal. 

(5) If an isosceles triangle be so placed 
that the unequal side forms the base, as in 
Fig. 19, and a perpendicular to the base is drawn from the 
vertex of the angle opposite, the perpendicular divides the base 
into two equal parts. The perpendicular is then said to bisect the 
base, the word bisect meaning to cut in halves. In Fig. 19, AC = 
CB, and CD is perpendicular to AB; therefore, AD = DB. 

(6) If two angles of one triangle are equal to two angles of 
another triangle, the third angle of the first triangle is equal to 




52 ELEMENTARY APPLIED MATHEMATICS §2 

the third angle of the second, because the sum of the two angles 
subtracted from 180° is the same in both cases. In Fig. 20, if 
A = A'andC = C, then B = B'. 

64. Similar Triangles. — If the angles of one triangle are equal 
to the angles of another, the two triangles are said to be similar. 
If the sides of one triangle are equal to the sides of another, then 
the triangles are equal. 

When two triangles are similar, as ABC and A'B'C in Fig. 20, 
one may be superposed on the other; that is, the vertex of one of 





Fig. 20. 

the angles of one triangle may be placed over the vertex of the 
equal angle of the other triangle, and the sides of the two angles 
can then be made to coincide, since they both have the same direc- 
tion from the common vertex, see Fig. 21. Here the vertex 
C of the angle C in Fig. 20 is placed over C, and the sides CA 
and CB are made to coincide with the sides C'A' and C'B') then 
AB is parallel to A'B'. 

65. When two triangles are similar, the sides opposite the equal 
angles are proportional. Thus, in Fig. 20, if the triangles ABC 
and A'B'C are similar, and angle A = angle A' and angle C 
= angle C , then AB : A'B' = AC : A'C; also, AB : A'B' = BC: 
B'C, and BC : B'C = AC : A'C. This is a very important 
principle, and should be remembered. 

Had the angle A been superposed on the angle A', Fig. 20, the 
result would be as shown in Fig. 21, the dotted hne5C being paral- 
lel to B'C. That AB, Fig. 2 1 , is parallel to A'B' is evident from the 
fact that since angle A = angle A' and the sides AC and A'C 
coincide, AB has the same direction as A'B', and when two right 




§2 MENSURATION OF PLANE FIGURES 53 

lines have the same direction, they must either coincide or be 
parallel. For the same reason, BC is parallel to B'C. Hence, 

if two sides and an angle of one triangle are equal to two sides and 
an angle of another triangle, the triangles are equal. 

66. The Right Triangle. — The right triangle is so important 
that it requires special treatment. 
Referring to Fig. 22, ABC is a right 
triangle, right-angled at C. The 
side C, which is opposite the right 
angle, is called the hypotenuse; the 
hypotenuse is alwaj^s the longest side 
of a right triangle, since it is opposite 
the largest angle. The other two 
sides are called the short sides or 
legs. In every right triangle, the square of the hypotenuse is 
equal to the sum of the squares of the legs; that is, referring to 
Fig. 22, 

c^ = a^ -^ 62 (1) 

This is a very important principle, and should be memorized; 
it is used more than any other principle in mensuration. By 
means of it, any side of a right triangle can be found if the lengths 
of the other two are known. For example, if a and b are known, 

c = Va^ + fe' (2) 
If c and a are known, h = sj c"^ — a^ (3) 
If c and 6 are known, a = -sj c^ — 6^ (4) 

Formulas (2), (3), and (i) are derived from formula (1) by 
solving (1) for c, 6, and a. 

Example 1. — Referring to Fig. 22, suppose the length of AC is 4% in. 
and the length of BC is 2% in.; what is the length of AB1 

SoLTJTiON. — Since 4% = 4.375, and 2% = 2.875, substitute the values 
for a and 6 in formula (2), obtaining 

c = V4.3752 + 2.8752 = 5.2351 - in. Ans. 

It is evident that formula (1) might have been used instead of 
formula (2); in fact formula (1) may be used in every case of this 
kind. 

Example 2. — In Fig. 23, P and P' are two pulleys, whose centers are 10 
ft. 8K in. and 4 ft. 6% in., respectively, from the floor AB, which is sup- 
posed to be level. By dropping plumb hnes from the shafts, the horizontal 
distance between the pulley centers is found to be 14 ft. 734 in. What is 
the distance O'O between the pulley centers? 



54 



ELEMENTARY APPLIED MATHEMATICS 



§2 



Solution. — Since AB is horizontal and OC and O'D are plumb lines, the 
angle OCD is a right angle. Drawing O'E parallel to OCD, it must be per- 
pendicular to OC (since AB is perpendicular to OC), and O'EO is also a right 
angle. Drawing O'O, O'EO is a right triangle, in which the side O'E = 14 
ft. 7.25 in. and OE = 10 ft. 8.5 - 4 ft. 6.75 in. = 6 ft. 1.75 in. For con- 




FiG. 23. 

venience in calculation, reduce the lengths of these sides to inches, obtain- 
ing O'E = 175.25 in. and OE = 73.75 in. Substituting these values in 
formula (2), O'O = V175.252 + 73.752 = 190.14- in. = 15 ft. 10}i in., 
very nearly (0.14 = }i nearly). Ans. 

67. Equality of Angles. — In connection with the last example, 
it was stated that the angle O'EO was a right angle, because 
O'E was perpendicular to OC. It might have been stated that 
O'EO is a right angle because it is equal to OCD, which is a right 




Fig. 24. 

angle, in accordance with the following principle: // the two 
sides of one angle are parallel to the two sides of another angle, 
both sides of which extend in the same or both in opposite 
directions, the two angles are equal. Thus, referring to Fig. 24, if 
A'O' be parallel to AO and B'O' be parallel to BO, both sides 
extending in the same direction, A'O'B' is equal to AOB. Again, 
if A"0" be parallel to AO and B"0" be parallel to BO, both sides 



§2 



MENSURATION OF PLANE FIGURES 



55 



extending in opposite directions to ^0 and BO, A"0"B" is equal 
to AOB. If one side of an angle coincide with a side of another 
angle, and the second side of the first angle be parallel to the second 
side of the other angle, the two angles are equal. Thus, if 0'" 
B'" and 0""B"" be parallel to OB, Fig. 24, AO"'B"' and 00'" 
B'" are equal to AOB. This a special case of the above principle, 
in which two of the parallel hues coincide. For the same reason 
rOB" = 00""B"" = AOB. Referring to Fig. 23, OE coincides 
with OC, and since O'E is parallel to CD, O'EO = DCO. 




,'''0 




A-' 



Fig. 25. 

A second principle applying to equahty of angles is: If the 
two sides of one angle are perpendicular to the two sides of another 
angle and both angles are acute or both obtuse, the two angles are 
equal. Referring to Fig. 25, suppose that A'O' and B'O' are per- 
pendicular to AO and BO, respectively; then AVB' = AOB. 
Again, suppose that 0"B" and 0"A" are perpendicular to OB 
and OA, respectively; then B"0"A" = BOA. Finally, suppose 
that A"'0"' and B'"0"' are respectively perpendicular to ^0 
and BO produced (indicated by the dotted hues; then, by the 
first principle, A""OB"" = AOB, and A"V"'B'" = AOB. 

These two principles are very important and are frequently 
used in connection with practical problems. 

68. Area of Triangles.— In Fig. 26, let ^C be the base of 
the triangle ABC. From the vertex of the angle B opposite the 
base, drop a perpendicular BD; the point where the perpendicular 
cuts the base is called the foot of the perpendicular, and that part 



56 



ELEMENTARY APPLIED MATHEMATICS 



§2 



of it included between the vertex B and the foot D is called the 
altitude of the triangle. The word altitude means height, and 
it is usually denoted in formulas by the letter h. In Fig. 27, 
it is necessary to produce the base in order that it may be cut by 
a perpendicular from B. In both figures, h = BD is the altitude. 




C A 




In any triangle, the area is equal to half the product of the base 
and altitude. Hence, letting A = the area, h = the base, and 
h = the altitude. 

Any side may be taken as the base, the altitude being the 
perpendicular distance between the base and the vertex of the 
angle opposite the base. Thus, in Fig. 28, 

AC XBD ^ BC XAF ^ AB X CE 
2 ~ 2 2 * 

Here AF is the altitude when BC is the base, and CE is the 
altitude when AE is the base. 

In a right triangle, if one leg is taken as the base, the other leg 

is the altitude, see Fig. 22. 
j„- /^\bx'^. j^ Consequently, the area of 

a right triangle is equal to 

half the product of its legs. 

If two triangles have the 

same base and the same 

altitude, their areas are 

equal. Thus, in Fig. 28, 

if MN is paraUel to AC, 

the altitudes of the three triangles ABC, ^ilfC, and ylA^C are equal; 

and since they all have the same base A C, all three triangles have 

the same area. 

Example. — If the base of a triangle is 2 ft. 9 in. and the altitude is 1 ft. 
4 in., what is the area of the triangle? 




§2 MENSURATION OF PLANE FIGURES 57 

Solution. — The lengths must be expressed in the same single unit, ex- 
pressing them in feet, 2 ft. 9 in. = 2.75 = 2f ft.; 1 ft. 4 in. = 1^ ft. 
Then, applying the formula, A = (2.75 X 1§) -h 2 = 1.8| sq. ft. Ans. 

Or, expressing the lengths in inches, 2 ft. 9 in. = 33 in.; 1 ft. 4 in. 

= 16 in. Then, A = ^^^^ = 264 sq. in. = ^ = If - 1.8^ sq. ft. Ans. 

Always remember that feet multiplied by feet give square feet 
and inches multiplied by inches give square inches. 

69. If the triangle is isosceles (an equilateral triangle is also 
isosceles), and the unequal side is taken as the base, the line repre- 
senting the altitude bisects the base, see (5) of Art. 63. Hence, 
referring to Fig. 19, if the three sides a, h, and c are known, a 

and 6 being equal, the altitude CD = h = -v &^ — (2) 

= |\/462 — c2, by formula (4) or (3), Art. 66; and the area of the 
triangle is, 

A = I X c X iV462 - c2 = |\/462 - c\ 

Example. — In an equilateral triangle, the length of each side is 73^ in. ; 

what is the area of the triangle? 

7 25 ■ 

Solution.— In this case, & = c; hence, A = ^V 4 X 7.25" - 7.25^ 

= -^VB X 7.252 = ^^ V3 = 22.760+ sq. m. Ans. 

For purposes of reference, let a = one of the equal sides of an 
isosceles triangle, and let c be the unequal side ; then 

A = ^V4:a^- c2 (1) 

For an equilateral triangle, let c = one of the sides; then 

A = ^Vs = .43301c2 (2) 

70. If all three sides of any triangle are known and the altitude 
is not known and cannot conveniently be measured, let p = <x 
+ 6 + c; that is, p equals the sum of the three sides. Then, 

A = i\p{v - 2a) (p - 2b) (p - 2c) 
In this formula, a, b, and c, are the three sides of the triangle. 

Example. — Suppose the three sides of a triangle are 6| in., 5f in., and 8| 
in.; what is the area of the triangle? 

Solution. — Here p = 6.5 + 5.75 + 8.125 = 20.375. It does not matter 
which of the sides are designated by a, by b, or by c; hence, taking them in 
the order given, p - 2a = 20.375 - 2 X 6.5 = 7.375, p - 2& = 20.375 



58 



ELEMENTARY APPLIED MATHEMATICS 



§2 



- 2 X 5.75 = 8.875, and p - 2c = 20.375 - 2 X 8.125 = 4.125. Substi- 
tuting these values in the formula, 

A =i\/20.375 X 7.375 X 8.875 X 4.125 = 18.542+. Ans. 

When the triangle is equilateral, a = b = c, p = 3a, and A 

= I \3a{3a - 2a) (Sa - 2a) (3a - 2a) = i \l3a' = ^ ^3, which 

is the same as formula (2), Art. 69, when a is substituted for c. 
Evidently p is the distance around the triangle; it is called the 
perimeter, peri meaning around and meter meaning measure. 
The word is applied to the distance around any plane figure ; and 
when the plane figure is a polygon, the perimeter of the polygon 
is the sum of the lengths of its sides. 

71. Projections. — Hereafter, unless a curve or broken Hne is 
specified, the word line will be understood to mean a right line — 
usually, a right-line segment. 

If from any point a perpendicular be let fall upon a line, the 
point in which the perpendicular intersects the line (the foot of 




the perpendicular) is called the projection of the point upon the 
line; and the distance from the point to its projection is the per- 
pendicular distance between the point and the line. Thus, in 
Fig. 29, A', B , C, and D' are the projections oi A, B, C, and D 
upon the Hne MN. 

The perpendicular distance from a point to a line is always the 
shortest distance from the point to the Hne. Thus, if from B, 
any Hne BE be drawn, BE is the hypotenuse of a right triangle 
of which BB', the perpendicular distance, is one leg, and the 
hypotenuse is always the longest side of a right triangle. 

If two points of any line are projected on another line in the 
same plane, the line segment between the two points of projection 



§2 



MENSURATION OF PLANE FIGURES 



59 



is called the projection of the line segment between the two points 
projected. Thus, in Fig. 29, A'B' is the projection of AB upon 
MN, and CD' is the projection of CD upon MN. Hence, to 
project any line, straight or curved, upon a right line in the same 
plane, project its two end points upon the right line, and the 
segment included between the two points of projection will be the 
projection of the given line. In Fig. 30, A'B', CD', and E'F' 
are the projections of AB, CD, and EF upon the line MN. 

C D E^ 



M- 




4' 



B' 



m 



E' 



F' 



-y 



Fig. 30. 



It will be noted that the projection of a line is always shorter 
than the given Hne, except when the given line is a right line 
parallel to the line of projection; in which case, the given line 
and the projected line are equal. In Fig. 30, AB is parallel to 
MN; hence, A'B' = AB. 

72. Relation between Sides of any Triangle. — In Figs. 31 
and 32, suppose that the angle A is acute. Drop a perpendicular 




from B to AC; then AD is the projection of AB upon AC. Rep- 
resent the projection AD by p. In Fig. 31, DC = h — p, and 
in the right triangle BDC, BD^ = a^ - {b - p)^ = a^ - b^ -{- 
2bp — p\ In the right triangle, BDA, BD^ = c^ — p^; therefore, 
BD^ = a^ — fe^ + 2bp — p^ = c^ — p^, or transposing terms, 
a2 ^ 52 ^ c2 _ 2bp (1) 



60 ELEMENTARY APPLIED MATHEMATICS §2 

In Fig. 32, DC = p - h, and in the right triangle BDC, 
BD^ = a2 - {p - h)^ = a^ - p^ i- 2ph - hK In the right 
triangle \BI>A, BD^ = c^ - p^; therefore, BD^ = a^ - p^ 
+ 2hp - &2 = c2 - p2. froni which, a^ = b^ + c^ - 2bp, which is 
the same as (1). 

Stated in words, the square of the side opposite an acute angle 
of any triangle is equal to the sum of the squares of the other two 

sides minus twice the product 

^^\ obtained by multiplying one of 

^^^^ / I these sides by the projection of 

S^"^^^ / I the other side upon that side. 

\^^ /" ! Thus, in Fig. 31, the projec- 

^'-^ \y~^ — H tion of AC upon AB is AD' 

'^<- J -Afe^ ^ = p\ and a2 = 62 + c2 - 2cp'. 

\^ h / This, evidently, is the same 

^'^N^ y / case as Fig. 32, when the tri- 

-^jy angle is turned over and AB 

Pj^j_ 33 of Fig. 31 is made the base. 

In Fig. 33, DA = p-\-b,p 

being the projection DC of a upon AC. Then, BD"^ = a^ — p^ 

= c^ — (& + p)^ = c2 — 62 _ 2bp —p^; from which, 

c2 = a2 + 62 + 262> (2) 
Stating formula (2) in words, the square of the side opposite an 
obtuse angle of any triangle is equal to the sum of the squares of the 
other two sides plus twice the product obtained by multiplying one 
of these sides by the projection of the other side upon hat side. Thus, 
in Fig. 33, the projection of 6 upon BC is CD' = p', and c^ = a^ 
+ 62 + 2ap'. 

Example. — In the triangle ABC, in which the side A 5 is the longest, 
AB = 22 in., AC = 13 in., and the projection of AC upon AB measures 
9.73 in., what is the length of the side CBl 

Solution. — The sides AB and AC evidently include an acute angle, 
since the longest side is always opposite the largest angle; hence, formula 
(1) applies to the case, substituting c for h, so that the formula becomes 
a2 = ^2 -f- c2 — 2cp; whence. 



a = ^h^ + c2 - 2cp = V132 + 222 - 2 X 22 X 9.73 

= AB -= 15.029 in. Ans. 

If the triangle is a right triangle, the projection of one leg upon 
the other is a point, which has no length and is therefore 0. In 
this case, c^ = a2 + 62 + 26 X = a^ + 6^, which is the same 
as formula (1) in Art. 66. 



§2 



MENSURATION OF PLANE FIGURES 



61 



QUADRILATERALS 

73. The Parallelogram. — When the opposite sides of a quadri- 
lateral are parallel, the quadrilateral is called a parallelogram; 
Fig. 34 shows two parallelograms ABDC, the sides AB and CD 
being parallel and equal, and the sides AC and BD being also 
parallel and equal. 




When the interior angles are not all equal and the sides are not 
all equal the parallelogram is called a rhomboid; both parallelo- 
grams in Fig. 34 are rhomboids. 

If the angles are not all equal, but the sides are equal, the paral- 
lelogram is called a rhombus; see Fig. 35, in which AB = BD 
= DC = CA. 

The diagonal of a parallelogram is a hne drawn through the 
figure from the vertex of one acute angle to the vertex of 
the other acute angle, the line CB in Figs. 34 and 35; this is called 
the long diagonal. A hne AD, drawn from the vertex of one ob- 
tuse angle to the vertex of the other obtuse angle, Figs. 34 and 
35, is called the short diagonal. 

The perpendicular distance BE between two parallel sides, in 
Figs. 34 and 35, is called the 
altitude of the parallelogram. 

74. Some Properties of Paral- 
lelograms.— (1) The diagonally 
opposite 'angles of any parallelo- 
gram are equal; thus, in Figs. 34 
and 35,C = B and A = D. This 
is a consequence of Art. 67, since 
AB is parallel to CD and AC is parallel to BD. 

(2) A diagonal divides the parallelogram into two equal tri- 
angles. Thus, ACB = CDB and ACD = ABD. This is evident 
since the three sides of one triangle are equal to the three sides 
of the other, the diagonal being a common side. 

(3) The diagonals of a parallelogram bisect each other; that 




Fig. 35. 



62 



ELEMENTARY APPLIED MATHEMATICS 



§2 



is, P being the point of intersection of the diagonals, PA = PD 
and PB = PC. 

(4) The sum of the interior angles of any parallelogram is 
equal to 4 right angles or 360°. For, referring to Figs. 34 and 
35, angle BDE = angle ACD, and CDB + BDE = CDB + ACD 
= C -\- D = 2 right angles = 180°. Since A = D and B = C, 
A+B + D-\-C ^2 X 180° = 360°. Therefore, if one angle 
of a parallelogram is known, all are known. Thus, suppose the 
angle C in Fig. 35 is 55°; then angle D is 180° - 55° = 125°, 
A = 125°, and B = 55°. 

75. The side on which the parallelogram is supposed to stand 
is called the base; in Figs. 34 and 35, CD is the base. 

The area of any parallelogram is equal to the product of the 
base and altitude; thus, the areas of the parallelograms in Figs, 
34 and 35 is equal to CD X BE. Let A = the area, I = the 
length of the base, and h = the altitude; then, 

A = Ih. 

This follows at once from the rule for finding the area of a 
triangle. Thus, area of triangle CDB, Figs. 34 and 35, equals 
}i X BE X CD, and since CAB = CDB, the area of the paral- 
lelogram is2 X}iCD X BE = CD X BE. 



R 




c 




A 


(a) 


D 



A 

Fig. 36. 




76. If one angle of a parallelogram is a right angle, all the 
angles are right angles, in accordance with Art. 74, and the paral- 
lelogram is then called a rectangle; see (a). Fig. 36. Here A 
= B = C = D = a right angle = 90°, and BC = AD and AB = 
DC. 

If the four sides of a rectangle are equal, it is called a square. 
Thus, (6), Fig. 36, is a square, because AB = BC = CD = DA 
and A=B = C = D^a right angle. 



§2 



MENSURATION OF PLANE FIGURES 



63 



Referring to (a), Fig. 36, it will be noted that if AD is the base, 
CD ■= AB is the altitude, and if AB is the base, AD = BC is 
the altitude. Hence, if one side of a rectangle be called the length, 
and one of the sides perpendicular to it be called the breadth or 
depth (according to whether the plane of the rectangle is sup- 
posed to be horizontal or vertical), the area of the rectangle is 
equal to the product of the length and breadth or depth; that is, 

A = lb = Id. 
in which A = the area, I = AD, Siudb = d = AB = DC. 

When the rectangle is a square, I = b = d, and A = I X I = l^. 
77. Referring to Fig. 37, ABCD is a rhombus. Draw the two 
diagonals, and represent the long diagonal 
by d and the short diagonal by d'. The 
diagonals intersect in P, which is the middle 
point of BD and AC (see Art. 74), and the 
four angles about P are right angles. That 
BPA = BPC is a right triangle is evident 
from (5) , Art. 63. Here ABC is an isosceles 
triangle and since P is the middle point of 
the base, the line from P to 5 is perpen- 
dicular to the base (see Fig. 19). Con- 
sidering the triangle DAB {= DCB), AP 

is the altitude and DB is the base. But 

d' 
AP = ^ and DB = d; hence, area oi DAB = }4 x d X %, and 

area of rhombus ABCD is 




k 



Fig. 37. 



d' 



A=\XdX~X2 = Ud' 



(1) 




sides 



Fig. 38. 

then, d"^ 



That is, the area of a rhombus is equal to 
one-half the product of its diagonals. 

Referring to Fig. 38, ABCD is a square— 
a rhombus whose angles are right angles. 
The diagonals AC and BD are equal and 
formula (1) becomes 

A =\dd = ^d\ (2) 

If it is desired to find the length of the 
diagonal of any square, let I = one of the 
-- l^ -\- l^ = 21', and 

d = VW = IV2 = 1.4142Z. (3) 



64 



ELEMENTARY APPLIED MATHEMATICS 



§2 



If the diagonal of a square is given and it is desired to find the 
length of the sides, l^ -\- l^ = d^ or 21^ = d^, and P = -^j 

from which I = ^^2 = .lOllOld (4) 

Example 1. — Fig. 37 is a drawing of a flat, rectangular plate having a 
hole in it shaped like a rhombus. From the dimensions given, find the 
area of the plate. 

Solution. — -The area of the plate is evidently equal to the area of the 
rectangle minus the area of the rhombus. 

Area of rectangle = 18.5 X 28.625 = 529.5625 sq. in. 

Area of rhombus = M X 10.375 X 12.75 = 66.1406 sq. in. 

area of plate = 463.4219 sq. in. 

Therefore, area of plate is 463.92 sq. in. Ans. 

Example 2. — What is the diagonal of a square, one side of which measures 
9^6 in.? 

Solution. — Substituting in formula (3), 

d = 1.4142 X 9%6 = 13.170- in. Ans. 
Example 3. — If the diagonal of a square is 11.82 in., what is the length of 
one of the sides? 

Solution. — Applying formula (4), 

I = .707107 X 11.82 = 8.358 in. Ans. 

78. Other Quadrilaterals. — If two sides of a quadrilateral are 
parallel and the other two sides are not parallel, the quadrilateral 
is called a trapezoid. In Fig. 39, {a) and (5) are trapezoids, the 




rb) 



n 



side BC being parallel to AD. The trapezoid in (6) is peculiar 
from the fact that it has two right angles, situated at A and B. 

If no two sides of a quadrilateral are parallel, it is then called 
a trapezium ; see Fig. 40. 

To find the area of a trapezoid, divide it into two triangles by- 
drawing a diagonal AC; then, if BC be taken as the base of one 
triangle and ^Z> as the base of the other, the altitude of these 
triangles is^D = CE in (a). Fig. 39, and AB = the perpendicular 
distance between the parallel sides in (b), Fig. 39. Denoting the 



§2 MENSURATION OF PLANE FIGURES 65 

altitude by h, the side BC by a, and the side AD by h, the area 
of one triangle is }-'2ah; of the other, 3^^6/i; and the area of the 
trapezoid is }i ah + K 6/i = }'2h{a + 6); or, 

This is a very important formula; it is frequently used, and 
should be carefully committed to memory. Stated in words, the 
formula becomes the following rule : The area of a trapezoid is 




FiQ. 40. 

equal to half the sum of the parallel sides multiplied by the perpen- 
dicular distance between them. 

Example. — What is the area of a trapezoid, the lengths of the parallel 

sides being 6}4 in. and 7M in. and the altitude being 9% in.? 

Solution. — The sum of the parallel sides is 6.5 + 7.25 = 13.75; hence, 

^, . 13.75 X 9.375 „, ^„^ . . 
the area is ^ = 64.453+ sq. in. Ans. 

It is better not to divide by 2 until after the multiplication has 
been performed, unless one of the factors is an even number. In 
the example just given, both factors were odd numbers. 

79. To find the area of a trapezium, draw a diagonal, as BD, 
Fig. 40; thus dividing the trapezium into two triangles. Using 
the diagonal as a base, draw the perpendiculars AD and CE 
from the vertexes opposite the base; then, area of trapezium is 
equal to ^iBD X EC ^ HBD X AD = ^BD (AD + EC). 

If through one of the vertexes, say C, a line be drawn parallel 
to the base (diagonal) BD and a perpendicular be drawn to this 
line from the other vertex, the length of this perpendicular is 



66 ELEMENTARY APPLIED MATHEMATICS §2 

equal to AD + EC. Representing this perpendicular by h, 
the base (diagonal) by b, and the area by A , 

A = y2bh. 

80. The rules and formulas just given for finding the areas 
of quadrilaterals apply only to what are termed convex polygons. 
All polygons that have been illustrated up to this point are 
convex polygons; and in any convex polygon, if any one of their 
sides be produced at either end of the side, the produced part will 
lie outside of the bounding line of the polygon. In Fig. 41, (a) 
is a quadrilateral and (&) is a pentagon. If the sides AB or 



Fig. 41. 

CD be produced from the end B, they wiU enter the space included 
by the bounding Hnes of the polygons. Angles like ABC are 
called re-entrant angles, and polygons having one or more re- 
entrant angles are called concave or re-entrant polygons. 

To find the area of a re-entrant polygon, divide it into triangles, 
as indicated by the dotted lines, find the area of each triangle, 
and their sum will be the area of the polygon. 

Unless otherwise stated, all polygons are supposed to be convex 
polygons. 



REGULAR POLYGONS 

81. Except in the case of irregular figures bounded by right 
lines, most of the polygons that occur in practice are regular 
polygons. 

If, in any regular polygon having an even number of sides, a line 
(diagonal) be drawn from any vertex to the vertex opposite that 
is farthest away, the line will pass through what is called the 
geometrical center of the polygon; and if two such lines be drawn 



§2 



MENSURATION OF PLANE FIGURES 



67 



from different vertexes, they will intersect in the geometrical 
center. Thus, in Fig. 42, which represents a hexagon (a polygon 
with an even number of sides), AD, BE, and CF all intersect in 0, 
which is the geometrical center, or, more simply, the center, 
of the hexagon. 

If a regular polygon have an odd number of sides, as the pen- 
tagon. Fig. 43, and a perpendicular be drawn from any vertex 
to the side opposite, it will pass through the geometrical center 
of the polygon; and any two such perpendiculars will intersect 
in the center. Thus, AP, BQ, CR, DS, and ET, which are per- 
pendicular respectively to CD DE, EA, AB, and BC, all inter- 
sect in 0, the center of the pentagon. 



CPU 




The perpendicular from the center to one of the sides, as OP 
in Figs. 42 and 43, bisects the side; that is, PC = PD. This 
perpendicular is called the apothem. 

The lines AD, BE, etc.. Fig. 42, and AP, BQ, etc.. Fig. 43, 
divide the polygons into as many equal triangles as the poly- 
gons have sides and the sum of the areas of these triangles 
equals the areas of the polygons. The area of one triangle is, 
letting I = length of one side and a = the apothem, ^^ial. 
If n = the number of sides of the polygon, the area, A, of the 
polygon is 3^^ anl. But nl = the perimeter of the polygon = p; 
hence, 

A = }ipa. (1) 

Stated in words, the area of any regular polygon is equal to half its 
perimeter multiplied by its apothem. 



68 ELEMENTARY APPLIED MATHEMATICS §2 

TABLE OF REGULAR POLYGONS 



Number of 


Apothem 


Area 


Number of 


Apothem 


Area 


sides 


a 


k 


sides 


a 


k 


3 


0.28868 


0.43301 


15 


2.3523 


17.642 


4 


0.5 


1. 


16 


2.5137 


20.109 


5 


. 68819 


1 . 7205 


20 


3.1569 


31.569 


6 


0.86603 


2.5981 


24 


3.7979 


45.575 


7 


1.0383 


3.6339 


25 


3.9579 


49.474 


8 


1.2071 


.4.8284 


30 


4.7572 


71.358 


9 


1.3737 


6.1818 


32 


5.0766 


81.225 


10 


1.5388 


7.6942 


40 


6.3531 


127.06 


11 


1.7028 


9.3656 


48 


7.6285 


183.08 


12 


1.8660 


11.196 


64 


10.178 


325.69 



The area may also be found by means of the above table when 
the number of sides in the polygon is given in the table. To use 
the table, let k = the number in the column headed area that 
coresponds to the given number of sides; let I = the length of 
the given side; then, 

A = kl\ (2) 

For example suppose that the length of a side of a regular 
octagon is 33-^ in., and it is desired to find the area of the octagon. 
Referring to the table, when the number of sides is 8, fc = 4.8284; 
hence, the area is A = 4.8284 X 3.25^ = 51 sq. in. Ans. 

Example. — One side of a hexagonal bar of iron measures IH in- J what 
is the area of a cross section of the bar? 

Solution. — Referring to the table, when the number of sides is 6, k 
= 2.5981; hence, area = A = 2.5981 X (1^)^ = 7.3985-, say 7.398 sq. 
in. Ans. 

The apothem can be used to lay out the polygon ; the manner 
of doing this will be described later. The apothem as given in 
the table is for a side equal to 1 ; hence to find the actual length 
of the apothem when the length of a side of the polygon is given 
let I = length of side ; then actual length of apothem = la. In 
the last example, the actual length of the apothem is lii 
X .86603 = 1.4614 in. 



THE CIRCLE 

82. Definition. — The circle is a curve every point of which is 
equally distant from a point within it called the center. The 



§2 



MENSURATION OF PLANE FIGURES 



69 



curve shown in Fig. 44 is a circle, being the center. A circle 
may be described (drawn) in various ways. Thus, with a pin, 
punch two holes in a strip of heavy 
paper; put the pin through one 
hole and the point of a sharp pencil 
through the other hole; then, keep- 
ing the pin stationary, revolve the 
pencil about the pin, keeping the 
strip of paper stretched tight, and 
the pencil will describe a circle. 
This method is shown in Fig, 45. 
A better way is to use an instru- 
ment employed by draftsmen, 
called compasses. This instru- 
ment consists of two legs united at 
one end by a joint, which permits them to open and close to any 
desired distance apart. One leg has a needle point at one end, 




Fig. 44. 




Fig. 45. 



Fig. 46. 



and the other leg carries a pencil point or pen. By placing the 
needle point of the compasses at the center, the leg carrying the 



70 ELEMENTARY APPLIED MATHEMATICS §2 

pencil point or pen may be revolved about the needle point, 
thus describing the circle as shown in Fig. 46. 

83. Referring to Fig. 44, any part of a circle, as AD, ABC, etc. 
is called an arc of a circle, a circular arc, or simply, an arc; 
it is so called from its shape, the word arc meaning how. A 
right line joining the extremities of an arc is called the chord 
of the arc or, simply, the chord. Thus, ^C is the chord of the 
arc ABC, and DE is the chord of the arc DABCE. When 
the chord passes through the center of the circle, it divides the 
circle into two equal parts, each of which is called a semicircle, 
meaning half-circle, and the chord is then called a diameter of 
the circle or, simply, a diameter. In Fig. 44, DE is a diameter 
of the circle, because it passes through the center. For the same 
reason, BF is also a diameter. The arc DBE is equal to the arc 
DFE, and both are semicircles. The arc FDB is equal to the 
arc FEB, and both of these arcs are semicircles. 

A right line drawn from the center to the curve is called a 
radius of the circle ; thus, OD, Fig. 44, is a radius, and so is OA, 
OB, etc. The plural of radius is radii; hence, OD, OA, OB, OC, 
OE, and OF are radii, of the circle DABCEF. All radii of any 
circle are equal, by definition of the circle, since they equal the 
distance from the center to the curve. A radius is also equal to 
one-half the diameter, since the diameter DE = OE + OD, and 
OE = OD = the radius. Consequently, the diameter equals 
twice the radius. 

The perimeter of a circle is commonly called the circumference ; 
in geometry, it is called the periphery. The word periphery 
is applied to plane figures having curved outlines, while the word 
perimeter is applied to plane figures bounded by right lines. 

84. The word circle is also appHed to the area contained within 
the circumference, hence, by area of a circle, the area included by 
the circumference is always meant. The area included by an 
arc and two radii drawn to the extremities of the arc is called a 
sector; in Fig. 44, the area OABC is a sector of the circle DBEF. 
The area included between an arc and its chord is called a 
segment; the area ABC A is a segment of the circle DBEF. 

85. If a line be drawn from a point without a circle and is 
terminated by the circumference after passing through the cir- 
cle, such a line is called a secant. In Fig. 47, PA and PB are se- 
cants. Evidently, a secant intersects the circumference in two 



§2 



MENSURATION OF PLANE FIGURES 



71 




Fig. 47. 



points; thus, PA intersects the circumference in A and D, and 
PB intersects it in B and E. If, however, the secant just touches 
the circle, intersecting it in only one point, it is called a tangent ; 
thus, PC is a tangent, because it intersects the circle in only one 
point, the point C, which is called the 
point of tangency. PG is also a tangent, 
and F is the point of tangency. 

86. Some Properties of Circles.^ — (1) 
If a diameter he drawn perpendicular to 
any chord, it bisects the chord and also 
the arc. In Fig. 44, if BF is perpen- 
dicular to AC, AG = GC, and arc AB 
= arc BC. 

(2) Any angle whose vertex is the center 
of the circle is measured by the arc it inter- 
cepts; the word intercept here means the 

part of the circumference cut off by and included between the 
radii forming the sides of the angle. In Fig. 44, the angle COE is 
measured by the arc CE that is intercepted by the radii CO and 
CE. The angle AOC is measured by the intercepted arc ABC) 
the angle DO A , by the intercepted arc DA ; etc It is here under- 
stood that the circumference is supposed to be divided into de- 
grees, minutes, and seconds, as described in Art. 54. 

Angles whose vertexes are situated at the center of the circle 
are called central angles or angles at the center. 

(3) Since central angles are measured by the arcs they inter- 
cept, it is evident that a diameter perpendicular to the chord of 
the arc bisects the central angle that is measured by that arc. 
Thus, by (1), if the diameter BF is perpendicular to AC, arc 
AB = arc BC, and AOB = BOC, since both angles are measured 
by equal arcs. 

(4) If a right line be drawn perpendicular to any chord at its 
mdddle point, it will pass through the center of the circle having the 
same arc that is subtended by the chord. In Fig. 44, the arc ABC 
is subtended by the chord AC. If G is the middle point of the 
chord and BGF is perpendicular to AC, then BGF must pass 
through the center of the circle DBEF of which the arc ABC 
is a part. 

(5) Two circles are equal when the radius or diameter of one is 
equal to the radius or diameter of the other; two sectors are equal 
when the radius and chord of one are equal to the radius and chord 



72 



ELEMENTARY APPLIED MATHEMATICS 



§2 



of the other; and two segments or two arcs are equal when the radius 
and chord of one are equal to the radius and chord of the other. 

(6) // the vertex of an angle lies on the circumference, the angle 
is measured hy one-half the intercepted arc. In Fig. 48, the vertex 
of BAC lies on the circumference; it is therefore measured by one- 
half the arc BC. Angles whose vertexes lie on the circumference 





Fig. 48. 



Fig. 49. 



are called inscribed angles; hence, an inscribed angle is one-half 
as large as the central angle having the same arc. Thus, BAG 
= H BOC, and BOG = 2 BAG. 

(7) // the inscribed angle intercepts a semicircle, the angle is a 
right angle, since a semicircle contains 360° -f- 2 = 180°, and one- 
half of 180° is 90°, a right angle. Thus, in Fig. 49, if AG is a 
diameter, .A 5C, ADC, and AEG are all right angles. Hence, 




Fig. 50. 

any angle inscribed in a semicircle is a right angle. This fact is 
made use of by mechanics to test the roundness of a semicircular 
hole, as shown in Fig. 50. Here a square is laid across the edges, 
and is then rotated back and forth. If the sides of the square 
just touch the edges and the point of the square just touches the 
bottom, the surface touched is semicircular, since, as shown in 



§2 



MENSURATION OF PLANE FIGURES 



73 



the figure, ABC is a semicircle and the angle ABC inscribed in it 
is a right angle. 

(8) If two chords intersect in a point within a circle, as DE and 
FG, Fig. 48, which intersect in H, the angle GHE, which equals 
DHF, is measured by one-half the sum of the arcs intercepted by 
these equal angles; that is, GHE (or DHF) is measured by one- 
half of arc GE + arc DF. 

(9) If two secants are drawn from the same point, the angle 
between the secants is measured by one-half the difference of 
the arcs they intercept. In Fig. 47, APB is measured by one- 
half of arc AB — arc DE. 

(10) The angle between a secant and a tangent drawn from the 
same point, also the angle between two tangents drawn from the 
same point, is measured by one-half the difference of the inter- 
cepted arcs. In Fig. 47, the angle APF is measured by one-half 
of arc AF — arc FD, and 
angle FPC is measured by 
one-half arc of FBC — arc 
FEC. 

(11) The radius drawn to 
the point of tangency is per- 
pendicular to the tangent 
In Fig. 47, if C and F are 
points of tangency, OC is 
perpendicular to PC and 
OF is perpendicular to PF. 

(12) If two circles in- 
tersect, the line passing 
through their centers is 
perpendicular to their com- 
mon chord and bisects the 
chord. In Fig. 51, two 
circles having the centers 
O and 0' intersect; AB 
is their common chord, drawn through the points of intersec- 
tion; then, the right line passing through the centers and 0' 
is perpendicular to AB and bisects AB. The circles whose 
centers are and 0" also intersect, their common chord 
being CD. Then, the right hne passing through O and 0" is 
perpendicular to CD and bisects CD. In the first case, the center 
of the second circle is outside the circumference of the first; but, 




74 



ELEMENTARY APPLIED MATHEMATICS 



§2 



in the second case, the center of the second circle is within the 
circumference of the first. 

(13) From a given point without a circle, two tangents may be 
drawn to the circle, as PF and PC in Fig. 47. If from the given 
point, a secant be drawn passing through the center, it bisects the 
angle formed by the two tangents and is perpendicular to the 
chord joining the points of tangency. In Fig. 47, if F and C 
are the points of tangency of the tangents drawn from P, and 
PB is a secant passing through the center 0, then OPF = OPC = 
^^FPC, and PB bisects the chord FC. PB is also perpendicular to 
FC. It is evident, also, that PB bisects the central angle FOC. 

87. Three Important Principles. — (1) // any two chords of a 
circle intersect, the product of the segments of one line is equal to 
the product of the segments of the other line, the segments being 
determined by the point of intersection. In Fig. 52, the chords 
AB and CD intersect in the point M; then, AM X MB = CM 

X MD. The chords AB 
and JK intersect in the 
point N; then, AN X NB 
= JN X NK. 

If EF is a diameter per- 
pendicular to the chord 
GH, IG = IH, according to 
(1), Art. 86, and IE X IF 
= IGXIH = IG^ = IH^; 
that is, a diameter per- 
pendicular to a chord is 
divided by the chord into 
two segments whose pro- 
duct is equal to the square 
of half the chord. The 
segment included between the chord and the arc is called the 
height of the arc or height of the segment; hence if the chord 
and height of the arc are known, the diameter or radius can be 
found. For, let c = the chord GH, h = the height IE, and d = 

c I c\^ 

the diameter; then IF = d — h, IG = ^, and h (d — h) = i^j ; 

from which, hd — h^ = 




or 



d = 



4/i 



(1) 



§2 MENSURATION OF PLANE FIGURES 75 

Representing the radius by r, r = ^^ ^^'^ 

If r and c are known and it is desired to find h, it may be 
found from formula (2), its value being 

h = r + ^ -v/4r^ — c^ 
When h is less than r, the minus sign is used, and 

h = r - W^r^ - C" (3) 

When h is greater than r, the plus sign is used, and 

h =r + W^r^ - c^ (4) 

If r and h are given and c is desired, 

c = 2V(2r - h)h (5) 

(2) 7/ from a point without a circle two secants are drawn, the 
product of the whole secant and the external segment of one line is 
equal to the product of the whole 
secant and the external segment 
of the other line. In Fig. 53, 
PA and PB are secants drawn 
from the point P; EP and FP 
are the external segments; 
then, PA XPE = PB X PF. 

(3) If from a point without 
a circle a secant and a tangent 
are drawn, the product of the 
whole secant and its external 
segment is equal to the square 
of the tangent. In Fig. 53, let 
PA be any secant and PC a 
tangent, both drawn from P; 
then, PA X PE = PC\ In 
order to measure PC, it is 
necessary to know the point 
of tangency C; and this can 
be found by drawing from 

the center a perpendicular to PC, the point where it intersects 
PC being the point of tangency. 

Example .^ — ^The chord of an arc has a length of 14% in.; the height of the 
arc is 3)4 in.; what is the radius? 




Fig. 53. 



76 



ELEMENTARY APPLIED MATHEMATICS 



§2 



Solution. — Applying formula (2), c = 14.875, h 
14.8752 + 4 X 3.252 



3.25, and 
= 10.135+ in. Ans. 



8 X 3.25 

88. It is sometimes desirable to know the chord and height 
of half the arc when the chord and height of the whole arc are 
given. In such a case, formulas may be found as follows: 

Referring to Fig. 54, let C 
= AB3.ndH = CD, the chord 
and height of the arc ACB; 
let c = AC and h = EF, the 
chord and height of the arc 
AC = half the arc ACB; and 
let r = the radius of the arc 
ACB. In the right triangle 
ADC, AC^= AD^ + CD\ or 




Q 



-{■H' 



C^ + 4:H^ 

4 



Multiplying and dividing this 
fraction by 2H, which of 
course does not alter its value, 
2H(C^ + 4H^) 



the result is 



C^ + 4g2 
8H • 



= 2rH, since by formula (2), Art. 87, r = 

c2 = 2rH, and 

c = \/2rH (.1) 

To find h, EF XFH = AF^ or h{2r -h)= (|) 
FH = 2r- h. But c^ = 2rH; hence, h(2r - h) 



8H 
Therefore, 



2rH 
4 



4 
rH 



smce 



and 



2rh 



rH 



h^ = ^- Dividing both members of this equation by — 1, 



to change the sign of h}, 



h^ - 2rh = - 



rH 



Solving this equation by the regular rule for quadratics (Art. 



31) 



2r + 



h = 



V 



(2r)2 + 4 X 1 X-~ 



2 X 1 



= r±^r{r-^)^ 



= r + 



(2) 



Vr^ 






§2 MENSURATION OF PLANE FIGURES 77 

To apply either of these formulas, it is first necessary to calcu- 
late the radius. 

Example. — Referring to the example in Art. 87, find the chord and height 
of half the arc. 

Solution.— The chord AB = 14.875 in., the height CD = 3.25, from 
which the radius was found to be 10.136 in. By formula (1), the chord of 
half the arc = c = -\/2 X 10.135 X 3. 25 = 8.1165 -, say 8.116 in. Ans. 

By formula (2), h = 10.135 ± a/i0.135^ - ^^'^^^ ^ ^'^^ = 10.135 

+ 9.287 = 19.422 in.; or, 10.135 - 9.287 = .848 in. The smaller value is 
evidently the one required in this case; hence, h = .848 in. Ans. 

It may be remarked that the smaller of the two values just ob- 
tained is the length of EF, while the larger value is the length 
FH; the sum of these two lengths is 19.422 + .848 = 20.270 
= EH = 2r = 2 X 10.135 = 20.270. 

89. Circumference and Area of the Circle. — The circumfer- 
ence of a circle is equal to the diameter multiplied by a number 
that is universally represented by the Greek letter t (pronounced 
pi or pe). This number has been calculated to 707 decimal 
places, and it has been proven that it cannot be expressed by 
a finite number of figures; its value to 9 significant figures is 
3.14159265 + , and is usually expressed as 3.1416; for rough 
calculations, 3t = --j- is commonly used for tt. Letting c = the 
circumference, d = the diameter, and A = the area, 

c = ird = 3.1416 d (1) 

Smce the diameter equals twice the radius, 

c = 27rr (2) 

The area of a circle is equal to x times the square of the radius, 
or 

A =Trr^ (3) 



Since 



r = |, A = 7r(|) = i7rrf2 = .7854^2 (4) 



These four formulas are extremely important; they should be 
carefully committed to memory. 
From formula (1), 



<* - ; = due = -^'^'^ (^> 



From formula (2), 



r = if = .159155c (6) 



78 ELEMENTARY APPLIED MATHEMATICS §2 

From formula (3), 



From formula (5), 



= J^ = .56419 VA (7) 

d = J^ = 1.1284 VA (8) 

Formulas (5) to (8) may be used to calculate the radius or 
diameter when the circumference or area is known; as a rule, 
however, these formulas are not used, formulas (1) to (4) being 
preferred. 

Example 1. — A pulley has a diameter of 32 in. and makes 175 revolutions 
per minute; how fast does a point on the rim travel in feet per minute? 

Solution. — When the pulley has turned around once, the point will 
have traveled a distance equal to the circumference of the pulley, and since 
the pulley turns 175 times in one minute, the point will travel 175 times the 
circumference of the pulley in one minute. Consequently, the distance 
traveled by the point in one minute is ird X 175 = 3.1416 X 32 X 175 = 
17592.96 in. = 1466.08 ft., say 1466 ft. Hence, the speed of the pulley is 
1466 ft. per min. Ans. 

Example 2. — Suppose the speed of a belt is 3160 feet per minute and that 
it drives a pulley that makes 330 revolutions per minute; what is the diam- 
eter of the pulley? 

Solution. — The speed of a point on the circumference of the pulley is 

the same as the speed of the belt, assuming that there is no slipping of the 

belt. Consequently, as shown in example (1), 

ird X 330 = 3160 ft. = 37920 in., or 

37920 
d = „ „ = 36.5766 = 36f I in., very nearly. Ans. 

Example 3. — The piston of a steam engine has a diameter of 16 in.; 
what is the area of the piston surface touched by the steam? 

Solution. — The area touched by the steam is evidently the area of a 
circle having a diameter of 16 in. By formula (4), the area is 

A = .7854 X 162 = 201.0624, say 201 sq. in. Ans. 

Example .4. — It is desired to bore a hole that shall have an area of 10 sq. 
in.; what must be the diameter of the hole? 

Solution. — Either formula (4) or (8) may be used, but formula (8) is 
rather easier to apply; using it, therefore, 

d = 1.1284\/l0= 3.5683, say 3.568 in. Ans. 

Example 5. — The circumference of a flywheel was measured with a 
tape hne and found to be 35 ft, 103^ in. What is its diameter to the nearest 
one-eighth inch? 

Solution. — Either formula (1) or (5) may be used. If (1) be used, it 
will be necessary to divide the circumference by 3.1416, while if (5) be used, 
the circumference may be multiplied by .31831. Since most computers 



§2 



MENSURATION OF PLANE FIGURES 



79 



would rather multiply than divide, use formula (,5). Reducing the feet to 
inches, 35 ft. 10.25 in. = 430.25 in. Then, 

d = .31831 X 430.25 = 136.953 in. =11 ft. 5 in. to the nearest one- 
eighth inch. Ans. 

Example 6. — When no ambiguity (confusion) is likely, a circle may be 
designated by referring to its centers only. Fig. 55 shows two pulleys, 
O and O', driven by a belt. Suppose the diameter of O is 48 in. and it makes 
220 revolutions per minute. It is desired to have pulley 0' make 450 
revolutions per minute; what must be the diameter of 0'? 

Solution. — ^Let N and n be the number of revolutions per minute made 
by O and 0', respectively, being the larger and O' the smaller pulley; 
let D and d be the respective diameters of and 0'. The speed of the belt 




Fig. 66. 



in feet per minute is equal to the circumference of in feet multiplied by 
the number of times it turns in one minute; hence, if the diameters of the 

pulleys are given in inches, the speed of the belt is . The speed of the 

belt is also equal to the circumference in feet of O' multiplied by the number 



rdn 
l2" 



Therefore, 



tDN 
12 



of times it turns in one minute; that is, it equals 

= -^2". Dividing both members of this equation by yx, 

DN = dn (9) 

In words, the product of the diameter and number of revolutions of one 
pulley is equal to the product of the diameter and number of revolutions 
made in the same time by the other pulley. Provided the unit used to 
measure D and d is the same, it is immaterial what unit is used; that is, 
the diameters may be stated in feet, inches, millimeters, etc. 
Applying the formula to the present case, 

48 X 220 = d X 450 

48 X 220 
from which d = — -^ — - = 22.4| in. = 23H in. very nearly. 

The diameter of d must be in inches because the diameter of D is in 
inches, and both diameters must be measured in the same unit. Formula 
(9) is very important in connection with calculations pertaining to pulleys 
and belts, and should be carefully memorized. 



80 ELEMENTARY APPLIED MATHEMATICS §2 

90. Length of Open Belt. — When the belt passes over the 
pulleys without crossing the line O'O joining the centers, as in 
Fig. 55, it is called an open belt. Knowing the distance O'O 
between the centers and the diameters of the pulleys, it is fre- 
quently desired to know the length of the belt. There is no 
simple, exact formula that will give the length of the belt, but the 
formula given below is sufficiently exact for all practical purposes. 
The lines A' A and C'C are tangent to the pulleys (circles) 
and 0'. Drawing the radii OA and OC, A and C are the points 
of tangency. Drawing the radii O'A' and O'C, A' and C are 
points of tangency. Draw diameters BD and B'D' perpendicu- 
lar to O'O; then the angles AOB and COD are equal, since the 
tangents A' A and C'C must intersect in some point, say P (not 
shown here), and by (13) of Art. 86, arc AE = arc CE; but arc 
AB = 90° - arc AE = 90° - arc EC = DC, and AOB = COD. 
Since, O'B' is parallel to OB and O'A' is parahel to OA, B'O'A' 
= BOA. For the same reason, C'O'D' = COD, and all four 
angles are equal. The radii of the arcs B'A' and BA are not 
equal, and the length of the arc AB is not equal to the length 
of the arc A'B'. Let L = the length of the belt, and let C = the 
distance O'O between the centers; then it is evident that the 
length of the belt is L = 2 X A'A -\- semicircle BED + 2 X arc 
BA + semicircle B'F'D' - 2 X arc B'A'. The difficulty arises 
in finding an expression for the lengths of the arcs BA and B'A'. 
Using the same letters as before, the following formula is suf- 
ficiently exact for all practical purposes: 

L ^ 2C + |(i) + d) + ^^-^ (1) 

When using this formula, C, D, and d must all be measured in 
the same unit. If it is desired to measure L and C in feet and 
D and d in inches, the formula then reduces to 

L = 2C'- + .1309(1)" + d") + 576(f -^ (2) 

by substituting D" = j^ and d" = y^ for D and d, respectively. 

In formula (2), C means C feet, and D" and d" mean D inches, 
and d inches. 

Example. — What length of belt is required when the pulleys have diam- 
eters of 56 in. and 16 in. and the distance between the centers is 24 ft. 3 in.? 
Solution. — Substituting in formula (2) the value 24.25 = 24 ft. 3 in., 

(Ka _ 1 A')2 

for C;L =2 X 24.25 + .1309(56 + 16) + 57^ y. 24 25 " ^^.04- ft. Ans. 



§2 



MENSURATION OF PLANE FIGURES 



81 



Since .04 ft. = .48 in. say ]4 in., the length of the belt may be taken as 58 
ft. K in. 

91. Length of Crossed Belt. — When the belt passes over the 
pulleys so as to cross the line O'O joining the centers, as in Fig. 
56, it is called a crossed belt. A and Care the points of tangency 
for the pulley and A' and C are the points of tangency for 
the pulley 0' . As in the case of the open belt, the angles AOB, 
COD, A'O'B', and C'O'D' are all equal. The length of the belt 
is L = 2 X A'A + semicircle BFD + 2 X arc A5 + semicircle 




Fig. 56. 



B'F'D' + 2 X arc A'B'. As with the open belt, there is no 
simple formula giving an exact value for L, but the following is 
sufficiently exact for all practical purpose : 

{D + dY 



Or, 



L = 2C + ^(Z) + d) + 



L = 2C' + .1309(i)" + d") + 



4C 

{D" + d"Y 



(1) 



(2) 



576C' 

The letters in these formulas have the same values as in the 
formulas of Art. 90. It will be noted that the only difference 
between these formulas and those of Art. 90 is the sign of d in the 
last term. 

Example. — Using the same values as in the example of Art. 90, find the 
length of a crossed belt. 

SoiiUTiON. — Substituting the values given in formula (2), 

L = 2 X 24.25 + .1309(56 + 16) + 57^ x 24 25 " ^8.296- ft. = 58 ft. 
4^^ in., very nearly. Ans. It will be noted that the length of the crossed 
belt, in this case, is 4M - K = 4M in. longer than the open belt. 

92. Concentric and Eccentric Circles. — Two or more circles 
are said to be concentric when they have the same center. In 
Fig. 57, the circles ABC and A'B'C have the same center 0; 
they are, therefore, concentric circles. The distance A' A 



82 



ELEMENTARY APPLIED MATHEMATICS 



§2 



= B'B = C'C, each distance being that part of the radius included 
between the circles. 

If one circle lies within another, but does not have the same 
center, the circles are said to be eccentric. In Fig. 58, the center 
of the larger circle is 0, and the center of the smaller circle is 0' ; 
hence, these circles are eccentric circles. 

If, in Figs. 57 and 58, the small circle represents a hole, the 
area of the space included between the hole and the large circle 




Fig. 57. 



Fig. 58. 



is evidently equal to the area of the large circle less the area of the 
small circle. Let D = diameter of the large circle, usually called 
the outside diameter, and let d = the diameter of the small 
circle, usually called the inside diameter; then area of large 



circle is -[D^ and area of small circle is -rd^. 
4 4 



surface in either Fig. 57 or Fig. 58 is : 



A =^2)2 
4 



4 4^ 



d^). 



The area of the flat 



(1) 



If the radius is used instead of the diameter, 
A = t{R^ - r2), (2) 

in which R is the radius of the outer circle and r is the radius of 
the inner circle. 

Example. — A circular disk has a hole through it that is 73^ in. in diam- 
eter. The outside diameter is 103^ in. What is the area of the flat surface? 

Solution. — Applying formula (1), remembering that j = .7854. 

A = .7854 (10.25^ - 7.125^) = 42.645 -, say 42.64 sq. in. Ans. 



§2 



MENSURATION OF PLANE FIGURES 



83 



EXAMPLES 

(1) It is desired to bore a hole that shall have an area of exactly 2 sq. in.; 
what must be its diameter? Ans. 1.596- in. 

(2) The diameter of a rod is 4% in.; what is its circumference? 

Ans. 14|| in. 

(2) A pulley 19 in. in diameter and making 350 revolutions per minute 

drives a pulley 8 in. in diameter; how many revolutions per minute does the 

smaller pulley make? Ans. 831}^ r.p.m. 

(4) The outside diameter of a cross-section of a cyhnder is 27 in., the in- 
side diameter is 25 in.; what is the area of the cross-section? 

Ans. 81.68 + sq. in. 

(5) The circumference of a shaft is 22% in.; what is its diameter? 

Ans. 7% in. nearly. 

(6) Two pulleys having diameters of 80 in. and 20 in. are 22 ft. 6 in. be- 
tween centers; what length of open belt is required to connect these pulleys? 

Ans. 58 ft. 4i|V in. 

(7) Two pulleys that are 9 ft. 2 in. between centers are connected by a 
crossed belt; the diameters of the pulleys being 18 in. and 12 in., what is the 
length of the belt? Ans. 22 ft. 5i^ in. 



SECTORS AND SEGMENTS 

93. Circular Measure of Angles. — Instead of measuring an 
angle in degrees, minutes, and seconds, it may be measured by- 
means of its arc expressed in 
terms of its radius. For in- 
stance, referring to Fig. 59, 
angle AOB = angle DOE. 
The number of degrees in the 
angle DOE is to the number 
of degrees in the semicircle 
GDEH as arc DFE is to arc 
GFH. Letting v = angle 
DOE, v° : 180° = arc DFE : xr, 
r being the radius OD and irr 
being >^ X 27rr = length of arc 
of semicircle. From this pro- 




portion, v° = 
arc ACB 



180° X arc DFE 



irr 



= 57.296° X 



Fig. 59. 

arc DFE 



= 57.296' 



X 



R 



when R = radius OA of arc ACB. Now suppose 



that the arc DFE = the radius r, then the arc ACB must equal 
the radius R, and v° = 57.296° X ^ = 57.296° X ;^ = 57.296°; 



84 ELEMENTARY APPLIED MATHEMATICS §2 

that is, an arc of 57.296° is equal in length to the radius of the arc. 
Taking this arc as the unit of measure, in which ca,se, it is called a 

180° 
radian, a semicircle is equal to rn oq^o = 3.1416 =7r radians; a 

quadrant, or 90°, is half a semicircle and is equal to 3^^^ = ^ 

= 1.5708 radians; any other angle, as ABC, will be equal to the 
number of degrees in the angle divided by 57.296. An angle 
measured in this manner is said to be measured in radians or to 
be measured in circular measure. When an angle is expressed 
in degrees, it is said to be measured in angular measure. 

arc DFE 
From the equation v° = 57.296° , DFE being any arc 

and r its radius, 

arc DFE v° , ^^^ . ,. 

= -- „„^o = angle DFE m radians; 

r oY.zyb 

in other words, if the length of an arc he diijided hy its radius, the 
quotient will be the measure of the angle in radians. Let d (Greek 
letter, pronounced theta) be the angle in radians, let I = length 
of the arc, and r — the radius of the arc, then the last equation 
may be written 

^ = e (1) 

r 
From (1), I = rd (2) 

If an angle be expressed in radians, and it be desired to find its 
equivalent in angular measure, let v be the angle in degrees; 

y° 
then, smce ^^ ^qq" = ^' 

v° = 57.296°^. (3) 
If the angle is given in angular measure, and it is desired to 
express it in radians, 

^ = .0174533^° (4) 



57.296 

Example 1. — The length of an arc is 23% in. and the radius of the arc 
is 32 in.; what is the angle in radians, and what is the angle in angular 
measure? 

23 625 
Solution. — From formula (1), d = ' = .73828 radians. Ans. 

From formula (3), v = 57.296 X .73828 = 42.3005° = 42° 18' 1.8". Ans. 

Example 2. — A certain angle is equal to 21° 31' 26"; if the radius of an 
arc having this angle is 153^ in., what is the length of the arc? 

Solution. — 'Reducing the minutes and seconds to a decimal of a degree, 
21° 31' 26' = 21.5239°. By formula (4), 6 = .0174533 X 21.5239; by 
formula (,2), I = rd = 15.5 X .0174533 X 21.5239 = 5.823- in. Ans. 



§2 MENSURATION OF PLANE FIGURES 85 

94. Length of Circular Arc. — Referring to formula (1), Art. 
93, if r is equal to the unit of linear measure (1 foot, 1 inch, etc.), 

6 = ^ = I] in other words, the measure of an angle in radians 

is the length of the arc to a radius I. Thus, in Fig. 59, if OD = OE 
= 1 inch, and 6 is the circular measure of the angle AOB = v, 
the arc DFE = 6 inches. Further, if OA = r, the length of the 
arc ACB = r X 6 in. = rd in. 

As previously defined, the figure OACB is a sector. By means 
of the formulas and principles of Art. 93, the length of the arc 
ACB of any sector can be found when the radius and the central 
angle (in either angular or circular measure) are known. If the 
chord AB and the height CI of the arc ACB are given, there is no 
simple, exact formula for finding the length of the arc ACB, 
in such cases, the angle AOB would usually be found by means 
of a table of trigonometric functions; then, knowing the angle, the 
length I of the arc can be found the formula 

I = rd ^ .0174533ry (1) 

If, however, a table of trigonometric functions is not available 

or if it is not desired to use such a table, the following formula will 

give results sufficiently exact for all practical purposes 

^ 40r^(15 + 16^^) , . 

75 + 180^2 + 64^6 ^ ' 

1 . ,■ h height of arc , ,, ,. 

in which t = - = , ■ — 7 and r = the radius. 

c chord 01 arc 

To apply the formula, first calculate the value of ^; if r is not 
given, calculate it by formula (2), Art. 87. 

If the angle is not greater than a right angle, or 90°, the term 
64^^ may be omitted, and the formula then becomes 

, ^ 40rf(15 + le^'') ^ 8r^(15 + 16^^) , . 

When the angle is equal to 90°, - = t= ~ ^ = .207107; 

c ^ 

hence, if the value of t equals or exceeds .21, use formula (2); 

but if t is less than .21, formula (3) may be used. 

Example 1. — What is the length of an arc whose chord is 7 3^ in. and 

whose height is 2 H^ in.? 

a a- . 0,1 -r, 75 117 75 ,, 16 75 

SoLUTioN.-Smce f = 2H - 7A = 3^ - ^g- = 32 X m ^ 234 

= .320513 is greater than .21, use formula (2). By formula (2), Art. 87, 
„ (M)Ml4(2£i)2 
'^ ~ SX2U 



86 ELEMENTARY APPLIED MATHEMATICS §2 

Substituting in formula (2), 

, _ 40 X 4.02375 X .320513(15 + 16 X .320513^) ^ ^ .^» • 

75 + 180 X .3205132 + 64 X .320513^ ^- ' ' ^^- '^'^*- 

The correct value of Z to 5 significant figures is 9.1748 +in.; hence, the 
value as calculated is sufiiciently exact for practical purposes. 

Example 2. — If the radius of the arc is 54 in. and the height of the arc is 
8}4 in., what is the length of the arc and what is the central angle? 

Solution. — Referring to Fig. 59, suppose ACB is the given arc, and let 

OC be perpendicular to the chord AB; then AI = ^ and CI = h. In the 

right triangle AIO, 01 = r — h and OA = r. Therefore, ^ = V r^ — {r — h)^ 

= -^542 - (54 - 8.5)2 = 29.0818 in., and c = 29.0818 X 2 = 58.1636 

8 5 
in. Consequently, t = _„ " „ . = .14614. Since t is less than .21, formula 

(3) may be used, and 

, 8 X 54 X .14614(15 + 16 X .1 4614^) . 

^= 15 + 36 X .146142 = 61.422 m. Ans. 

The correct value of Z to 5 significant figures is 61.422 in. 

From formulas (1) and (3), Art. 93, = -, and z; = 57.2960 = 57.2-96- 

61 422 
= 57.296 X ^Ht- = 65.171°- = 65° 10' 15.6". Ans. 
04 

The exact value is 65° 10' 14.6"; hence, the error is only 1 second, which is 

too small to be considered in ordinary practice. 

If the angle is large, that is, if it is greater than 90°, and great 
accuracy is desired, it will be better to find the ratio of the chord 
and height of half the arc, which may be designated by t'; then 
substituting t' for t in formula (3), the length of half the arc will 
be found very closely. By means of formula (1), Art. 87, and 

formulas (1) and (2), Art. 88, the value of ^' = - for half the arc 
can be shown to be 

, _ VSr/t + c , .V 

^ 4l ^^^ 

Or, t' = ^^^'±^ (5) 

Having found t', the length of the whole arc will then be given 
by the formula, 

l^rf (15 + 16^^^) 

^ ~ 15 + 36«'2 y"^) 

Referring to the first of the two preceding examples, 

^ 4^2U ^ ^^^ 

Substituting this value of t' in formula (6), Z = 9.1749 + in. 



§2 MENSURATION OF PLANE FIGURES 87 

95. Area of Sector and Segment. — The area of any sector of a 
circle is equal to one-half the product of its radius and arc. Let 
I = the length of the arc and r = the radius, then 

A = >^rZ. (1) 

It will be noted that this formula is the same as for the area of a 
triangle when I = the base and r = the altitude. 
Since I = rd, substitute rd for I in (1), and 

A = yir^d (2) 

that is, the area of a sector is equal to one-half the product of the 
angle in radians and the square of the radius. 

Referring to Fig. 59, the segment ACBA = sector OACB 
- triangle OAB. Area of triangle OAB = }ixABX 01 ;hut AB 
is the chord of the arc ACB = c, and 01 = radius OC — minus 
height of arc = r — h. Consequently, area of segment is A 
= H^l - Hc(r - h), or 

. rl — c(r — h) r^d — c(r — h) .„, 

A = -^ ^ = 2 ^^) 

Another formula for finding the area of a segment (approxi- 
mately) is the following, in which D is the diameter; 

^ = f \/f - -^^ w 

This formula will give results sufficiently accurate for most 
practical purposes, and is rather easier to apply than (3), when 
it is necessary to calculate both the highest h and the angle d 
of the arc. 

Example. — Fig. 60 represents a round tank having an inside diameter 
of 60 in. The tank lies on a flat surface and is filled with water to a depth 
of 42 in. If the ends of the tank are flat and the tank is 12 ft. long, how 
many gallons of water are in the tank? 

Solution. — The volume of the water is equal to the area of the segment 
AGB multiplied by the length of the tank. To find the length of the arc 
AGB, calculate the length of the arc AFB and subtract it from the circum- 
ference of the circle AFBG. The height of the arc AFB is /i = 60 - 42 

= 18 in. To find the chord AB, apply principle (1), Art. 87, and (2)^ 

= jP^ X H(? = 18 X 42 = 756; from which, c = \/3024 = 54.991 in. Then, 

h 18 
since - = - . „„.. = .327326, use formula (2), Art. 94, and 

_ 40 X 30 X .327326(15 + 16 X .327326") ^ . 

75 + 180 X .3273262 + 64 X .327326« ^"* 

The circumference of the circle is 3.1416 X 60 = 188.496 in., and the 
length of the arc AGB is 188.496 - 69.573 = 118.923 in. The area of the 



88 



ELEMENTARY APPLIED MATHEMATICS 



segment AGB is evidently equal to the area of the sector AGBO plus the 
area of the triangle AOB; area of sector = i^ x 30 X 118. 923 = 1783.845 
sq. in.; area of triangle = M X 54.991 X (30 - 18) = 329.946 sq. in.; and 
area of segment = 1783.845 + 329.946 = 2113.791 sq. in. The length 
of the tank is 12 ft. = 144 in.; hence, the volume of the water is 2113.791 
X 144 = 304385.9 cu. in. Since one gallon contains 231 cu. in., the number 
of gallons in the tank is 304385.9 ^ 231 = 1317.7 -, say 1318 gallons. Ans. 




Fig. 60. 



The area of the segment might have been found by finding the area of the 
small segment AFB and subtracting it from the area of the circle. The 
flat surface ABCD represents the water level. 

96. Area of Fillet. — When two solids intersect so as to form a 
square corner, as shown in Fig. 61, the strength of the piece can 
be greatly increased by rounding the corner. This is usiially 

done by striking an arc of 
a circle having a small 
radius. The curved part 
thus added is called a fillet. 
While fillets add somewhat 
to the weight of the piece, 
it is customary to neglect 
this in the case of heavy 
castings; but if an accurate estimate of the weight is desired, 
it is necessary to include the weight of the fillets. Referring to 
Fig. 61, ACB is a fillet, and it is assumed that the sides AC and 
BC form a right angle. Letting OA = OB = r, the area of the 
fillet is equal to the area of the square AOBC — area of the 
quadrant AOB; hence, A ^ r^ - ^irr'^ = r^{l — .7854), or 

A = .2146r2. 





FiQ. 61. 



§2 MENSURATION OF PLANE FIGURES 89 

Example. — If the radius of a fillet is yi in,, what is the area of the fillet? 
Solution. — Applying the formula, A = .2146 X {^iY = .01341 sq. in. 

Ans. 

It will be noted that the area is quite small; and since the radius 
is never very large, it is sufficiently exact for practical purposes 
to take the value of A as one-fifth the square of the radius. 



EXAMPLES 

(1) If the length of an arc is 10.26 in. and its radius is 33.14 in., what is 
the angle in radians and also in angular measure? . f .30960— radians. 

^^s- \ 17° 44' 19/'. 

(2) If a certain angle measures 31° 12' 27", what is the length of the 
arc having this angle, the radius of the arc being 19.32 in.? Ans. 10.523 in. 

(3) The chord of an arc is 14^^ in., the height of the arc is 1% in., what is 
the length of the arc and what is the angle in radians and degrees? 

[ I = 15.378 in. 
Ans. \e= .99586 radian. 
\ v° = 57° 3' 32". 

(4) Referring to the example of Art. 95, how many gallons are in the 
tank when the depth of the water is 54 in.? Ans. 1671 gallons. 

(5) A fillet has a radius of %6 iii-j what is the area of the fillet? 

Ans. .021 — sq. in. 



INSCRIBED AND CIRCUMSCRIBED POLYGONS 

97. An inscribed polygon is one whose vertexes lie on the 
circumference of a circle. In Fig. 62 the vertexes of the poly- 
gon ABCDEF lie on the circumference of the circle ABCDEF, 
and it is, therefore, an inscribed polygon. A circumsciibed 
polygon is one whose sides are tangent to a circle. In Fig, 62. 
the sides of the polygon ABCDEF are all tangent to the circle 
A'B'C'D'E'F', and it is, therefore, a circumscribed polygon with 
reference to the circle A'B'C'D'E'F'. 

If the polygon is a regular polygon, it may always be inscribed 
in a circle, and it may also be circumscribed about a circle. The 
polygon in Fig. 62 is a regular hexagon, and it is a property of the 
regular hexagon that the sides are always equal in length to the 
radius of the circle in which the hexagon is inscribed; thus AB = 
BC = OC. Consequently, a regular hexagon may be drawn by 
describing a circle and then spacing off on its circumference 
chords equal in length to the radius. 

The center of the circles within which the regular polygon is 
inscribed and about which it is circumscribed is the geometrical 



90 



ELEMENTARY APPLIED MATHEMATICS 



§2 



a side, ^ = half a side= B'B, and (k)^= 



center of the polygon, and lines drawn from the center to the 
points of tangency, as OA', OB', etc. are apothems of the 
circumscribed polygon. Let a = the apothem and r = the 
radius of the circle within which the regular polygon is in- 
scribed; then, if s 
— a^, from which 

Given the radius r and the length s of the side, describe a circle 
with the radius r, and then space off on the circumference the 
length s. The value of a for polygons most commonly used may 
be obtained from the table given in Art. 81, in which a is a 
multiple of s. 




Fig. 62. 



Fig. 63. 



Example. — It is desired to inscribe a regular polygon of 11 sides in a circle 
having a diameter of 2J^ in. Show how this can be done by means of the 
table in Art. 81. 

Solution. — The apothems given in the table are multiples of the length 
of the side; hence, if the side be taken as 1 inch, the apothem for a regular 
polygon of 11 sides will be 1.7028 in., say 1.7 in. Draw two lines AB and 
OC at right angles to each other, Fig. 63. Make OC equal to 1.7 in., the 
apothem, and CB = CA = |s = | in., in this case (taking s as 1 in.), 
and draw OA and OB. With O as a center and a radius equal to the radius 
of the given circle = | X 2§ = 1| in., describe a circle; join the points 
E and F, the points of intersection of OA and OB with the circle, by the line 
EF, and EF will be one of the sides of the inscribed 11-sided polygon, and 
may be spaced off around the given circle. 



§2 



MENSURATION OF PLANE FIGURES 



91 



If it is desired to calculate the length of the side EF, first 

find the length of OB. In the right triangle OCB, CB = .5, 

and OC = 1.7028; hence, OB = V-S^ + 1.70282 - 1.7747 

in. The triangles AOB and EOF are similar; whence the pro- 

. EF OF s 1.25 
portion -Jd 



OB' ^^ 1 



and s = .70434 in. 



1.7747' 

If the line AB be taken as some length other than 1 inch, it 
must be multiplied by the value of the apothem as given in the 
table to find the length of OC. 



THE ELLIPSE 

98. The ellipse is a plain figure so constructed that the sum 
of the distances from any point on the curve to two fixed points 
is constant; thus, referring to Fig. 64, F and F' are the two fixed 
points, and CF + CF' ^ PF + PF' = QF -i- QF' = etc., and the 




closed curve is an elHpse. The longest line that can be drawn in 
the figure is the lineal 5, which passes through the two fixed points 
F and F'; it is called the major axis. From the center of the 
major axis, draw CD at right angles io AB; then CD is the short- 
est line that can be drawn in the ellipse, and is called the minor 
axis. The points A and B at the extremities of the major axis 
are called the vertexes of the ellipse; the point is called the 
center, and the fixed points F and F' are called the foci, either 



92 ELEMENTARY APPLIED MATHEMATICS §2 

point being called a focus of the ellipse. The vertexes, center, 
and foci all lie on the major axis, and the distance AF must 
equal the distance F'B; hence, for the point A, FA -\- F'A 
= F'A + F'B = AB; and the sum of the distances from any point 
on the curve to the foci is equal to the major axis. For the point 
C, CF = CF' = lAB = OA = OB. 

In practice, an ellipse is usually specified by giving the lengths 
of the major and minor axes, which are commonly called the long 
and short diameters. An ellipse may be drawn mechanically 
in the following manner: Lay off the long diameter (major 
axis), and bisect it, thus locating the center 0; through 0, draw 
a perpendicular CD, and lay off OC = OD = one-half the 
short diameter (minor axis); then, with C (or D) as a center 
and with a radius equal to one-half the long diameter, draw short 
arcs cutting AB in F and F', thus locating the foci. Stick pins 
in the paper at F and F' and also at C; tie one end of a piece of 
thread to one of the pins F orF', and pass the thread around the 
other two pins, drawing it taut and passing it several times around 
the pin at the other focus. Now pull out the pin at C, and with 
a pencil held perpendicular to the plane of the paper and pressing 
against the thread (but not hard enough to stretch it) , move it so 
that it keeps the thread tight, thus describing one-half of the el- 
lipse, say the upper half ACPB; then bring the thread to the 
other side of the pins, describe the other half of the eUipse or 
ADQB. 

99. Circumference and Area of Ellipse. — In works on mathe- 
matics, it is customary to denote one-half the major axis by the 
letter a and one-half the minor axis by the letter 6; thus, in Fig. 
64, a = OA = OB, and b = OC = OD. The area of an elhpse 
is 

A = irab (1) 

There is no exact formula giving the circumference (periphery) 

of an ellipse, but the following formula, in which r = — tj-t 

and p = the circumference (periphery) , gives results sufficiently 
exact for all practical purposes: 

V=Ha-\- b) ^^ ~ ^g^, (2) 

Example. — The long and short diameters of an ellipse are 163^ in. and 
4% in. respectively; what is the area of the ellipse? what is the circumference. 



§2 



MENSURATION OF PLANE FIGURES 



93 



Solution. — The area, by formula (1), is 

A - ,ai, - 3.1416 X ^« X f . ^« X 16.25 X 4.5 

= .7854 X 16.25 X 4.5 = 57.432+ sq. in. Ans. 
The circumference (periphery), by formula (2), is 



p = 3.1416 



16.25 , 4.5\ 64 - 3 X .56627^ 



/ 16.25 4^\ 64 - 
\ 2 "^ 2 / 64 - 



= 35.264 in. 



2 ' 2 / 64 - 16 X .566272 
To apply formula (2), first calculate the value of r; thus, r 
2a - 26 16.25 - 4.5 



Ans. 

a — b 



= .56627-. Then, r^ 



a + h 
,320662 -; r« = (r^)^ 



~ 2o + 26 16.25 + 4.5 

= .3206622 = .102824 + ; the remainder of the work is evident. 

It may be remarked that the major axis divides the ellipse 

into two equal parts; the minor axis also divides the eUipse into 

two equal parts. 

AREA OF ANY PLANE FIGURE 

100. If the figure can be divided into elementary plane figures 
{i.e. triangles, rectangles, circles, segments, etc.) the area of each 
of the elementary fig- 
ures may be calculated, 
and the sum will be the 
area of the entire figure. 
Referring to Fig. 65, the 
dotted lines show that 
the figure may be 
divided into three trape- 
zoids two of them equal 
— one rectangle, a seg- 
ment of a circle, and 
two equal fillets. From 
the dimensions marked 
on the drawing, the 
areas of all these ele- 
mentary figures may be 
found, and their sum 
will be the area of the 
entire figure. The work 
is as follows: 

There are two trapezoids of the same size as ahdc; the length of 
the side ac is 2^ + f = 3", and the area of the two trapezoids is 
3 + 2i 





/, 




B \h 




_: 






s" 


< \^" y 


< 3?" 


*- 


I. 
-• 


h ^& 


^'4*1 
C 


< -'a 




■ 


d 




-< 81 


b 

>- 





Fig. 65. 



A = 



X 31 X 2 = 17.72 sq. in. 



94 



ELEMENTARY APPLIED MATHEMATICS 



§2 



The length of the rectangle is 2| + f + 4| = 7i", and its 
area is 

^ = 7i X If = 12.47 sq. in. 
The length of the side ih of the trapezoid kjhi is If+lf+ll 



= 41" and its area is 



4f + 2i 



X 3 = 10.5 sq. in. 



. The area of the fillets is 2 X i X (f)^ = .06 sq. in. 

Concerning the segment, the radius and the chord are known and 
it is necessary to calculate the height. Using formula (3), 
Art. 87, h = r - ^W^r^ - c" = 1.5 - Ha/4 X 1.5^ - 2.252 
= .508". 

Substituting this value of h in formula (4), Art. 95, 



A = 



4 X 5082 2 X 1.5 



- .608 = .79 sq. in. 



3 \ .508 

The entire area is 17.72 + 12.47 + 10.5 + 0.06 + 0.79 
= 41.54 sq. in. 

101. If the figure is of such a nature that it cannot be divided 
into simple elementary figures, proceed as in Fig. 66. Draw a 




Fig. 66. 



line AB, either through the figure or outside of it, as in Fig. 66; 
then draw lines CB and DA tangent to the extreme ends of the 
given outline and perpendicular to the line AB. The intersec- 
tion of the tangents with AB locate the points A and B. Divide 
AB into any convenient even number of equal parts, in the present 
case 8, and through the points of division, erect perpendiculars 
(called ordinates) to AB; these intersect the figure in the lines 



§2 MENSURATION OF PLANE FIGURES 95 

a'a, h'h, c'c, etc., which are all parallel to one another, and are 
equally distant apart, li n = the number of parts into which 
AB has been divided, and h = the distance between two con- 

AB 
secutive ordinates, measured parallel to AB, h = -^• 

If, now, another series of ordinates are drawn midway between 
those previously drawn, here indicated by the dotted hnes, and 
the lengths of these dotted ordinates are measured and added, 
the sum so obtained multipHed by h will be the approximate area 
of the figure. This method is called the trapezoidal rule. The 
greater the number of parts into which AB is divided, that is, 
the smaller the value of h, the more accurate will be the result 
obtained for the area. 

102. Another rule that is more accurate than the method just 
described is what is known as Simpson's rule. Having drawn 
the ordinates dividing the figure into n equal parts (w being an 
even number), designate the ordinates by yo, yi, y^, etc., as shown 
in the figure. Then letting i/o and yn be the end ordinates, the 
area by Simpson's rule is 

A = I [?/o + i/n + 4(yi + 2/3 + 2/5 + etc.) + 2(i/2 + ?/4 + J/e + etc.)] 

Expressed in words, Simpson's rule is: 

Add the end ordinates, four times the odd ordinates, and 2 times 
the remaining even ordinates, and multiply this sum hy one-third 
the distance between the ordinates. 

Applying both methods to the outline given in Fig. 66, the 
lengths of the dotted ordinates, beginning with the left and pro- 
ceeding in order to the right, are 0.97", 1.52", 1.51", 1.04", 0.57", 
0.67", 0.85", and 0.75"; their sum is 7.88". The number of 
equal parts is 8 = n, and I = AB = 3.02". Therefore, h 

= ^-f, and A = ^-f X 7.88 ^^t^^.>f^ = 2.97 sq. in., by 

the trapezoidal rule. 

Applying Simpson's rule, ?/o and y^ are both equal to 0; yi 
= 1.34", yz = 1.34", y^ = 0.64", y^ = 0.85", and the sum of these 
odd ordinates is 4.17"; y^ = 1-56", y, = 0.67", y, = 0.83", 
and the sum of these even ordinates is 3.06". Therefore, by 

the rule, since ^ = -^— '^ 3 "^ .12583, 

A = .12583 (0 + + 4 X 4.17 + 2 X 3.06) = 2.87 sq. in. 



96 ELEMENTARY APPLIED MATHEMATICS §2 

103. It was stated in Art., 101 that the greater the number of 
parts into which AB is divided the greater the accuracy of the 
result, and this is true whichever method is used. If AB be 
divided into 16 equal parts ins^tead of 8, the area will be 2.954 
sq. in. by the trapezoidal rule and 2.890 by Simpson's rule. For 
32 equal divisions, Simpson's rule gives 2.919 sq. in. Tabulat- 
ing these results, 

Trapezoidal rule (8 parts) 2.975 sq. in. Error, +0.056 

Trapezoidal rule (16 parts) 2.954 sq. in. Error, +0.035 

Simpson's rule (8 parts) 2 . 869 sq. in. Error, — . 050 

Simpson's rule (16 parts) 2 . 890 sq. in. Error, — . 029 

Simpson 's rule (32 parts) 2 . 9 19 sq. in. Error, 

Assuming that the result obtained by Simpson's rule for 32 
parts is correct to all four figures, it will be observed that the 
error when using the trapezoidal rule is somewhat greater 
than when using Simpson 's rule. In practice, it is always advis- 
able to divide AB into at least 10 parts; this not only increases 
the accuracy, but it makes n a very convenient number to divide 
by when finding h. If particularly accurate results are desired, 
it is best to make n = 20, in which case, the result will be as 
accurate as the limits of measurement permit. 



ELEMENTARY APPLIED 
MATHEMATICS 

(PART 2) 



EXAMINATION QUESTIONS 

(1) The length of the diagonal of a square is 5i inches; what is 
(a) the length of one of the sides, and (6) what is the area? 

■ (a) 3.7123 in. 



^^^* I (h) 13.781 sq. in. 

(2) The lengths of the sides of a triangle are 29.1 ft., 21.8 ft., 
and 36.5 ft.; what is the area of the triangle? 

Ans. 317.18 sq. ft. 

(3) The side of a regular dodecagon measures 4| in.; what is 
the area of the polygon? Ans. 266.08 sq. in. 

(4) If (a) the diameter of a circle is 35f in., what is its area? 
(6) What must be the diameter of a circle that will enclose an 
area of 1800 sq. in.? .^i («) 1003.8 sq. in. 

^^^- I (6) 47.873 in. 

(5) The length of the chord of a circular arc is 46f in., the 
height of the arc is lOi in., what is (a) the radius? (6) the angle 
in radians? (c) the angle in degrees? First find length of arc. 

r (a) 31.636 in. 
Ans. \ (6) 1.6570 radians. 
I (c) 94° 56' 27" 

(6) Referring to the last example, what is (a) the chord of 
half the arc? (b) what is the height of half the arc? 

r (a) 25.466 in. 
^^^- 1 (&) 2.6756 in. 

(7) The diameters, of two belt pulleys are 64 in. and 14 in. 
and the distance between the centers is 18 ft. 9 in. (a) What 
length of open belt is required? (6) what length of crossed belt? 
Give lengths to nearest 16th of an inch. 

Ans /(«)47ft.ll^in. 
^''^- I (6) 48 ft. 3-i:V in. 
T 97 



98 



ELEMENTARY APPLIED MATHEMATICS 



§2 



(8) If the central angle of a circular arc is 121° 27' 36", (a) 
what is the angle in radians? (6) if the radius is 241 ft., what is 
the length of the arc? . f (a) 2.1199 radians. 

1(6) 51.938 ft. 

(9) Find from the dimensions given, the area of the outline 
shown in Fig. 1. The dotted lines indicate how the figure should 
be divided. As shown, the figure is divided into two semi-circles, 
two rectangles, two trapezoids, two sectors, and two fillets. 

Ans. 28.261 sq. in. 




Fig. 1. 



(10) If the chord of a circular arc is 8.5 in. and the height of 
the arc is li\ in., what is (a) the radius? (6) the length of the 
arc? (c) the angle in radians? (d) the angle in degrees? (e) the 



area of the sector? 



Ans. 



(a) 
(h) 
(c) 
id) 
(e) 



7.5372 in. 
9.0307 in. 
1.1981 radians. 
68° 38' 56" 
34.033 sq. in. 



§2 



EXAMINATION QUESTIONS 



99 



(11) Referring to the last question, (a) what is the area of the 
segment? (6) Calculate the area also by formula (4) of Art. 95. 

^'^^' 1(6) 7.575 sq. in. 

(12) If the major and minor axes of an ellipse are 37 in. and 
12| in., respectively, (a) what is the area? (6) what is the peri- 
phery of the elHpse? . / {a) 363.25 sq. in. 

^'^^- 1(6) 82.594 in. 

(13) The outhne shown in Fig. 2 resembles an indicator dia- 
gram showing the variation of pressure in a steam-engine cylin- 
der. Find the area of the diagram (a) by the trapezoidal rule, 



















f] 










\ 




y 


c 












1 

1 














D 






A 






iin r- 












jB 







Fig. 2. 

and also by (6) Simpson's rule. The end ordinates should be 
tangent to the curve at C and D and perpendicular to a hne 
parallel to AB. The value obtained by dividing the area by the 
perpendicular distance between the end ordinates is called the 
mean ordinate, and is indicated on the diagram by the dotted 
line MN; that is, it is equal to the perpendicular distance be- 
tween MN and AB. Find (c) value of the mean ordinate. Divide 
diagram into 12 equal parts; for the convenience of the student, 
the ordinates have been drawn in position. 

(14) A certain building lot has the shape of a sector of a circle 
whose radius is 149 ft., the frontage being 117 ft. What is the 
area of the lot, and what part of an acre is it? 

Ans. 8716.5 sq. ft. = .20010 A. 

(15) An opening is to have the shape of an equilateral triangle 
and is required to have an area of 16 sq. in.; what is the length 
of one of the sides? Ans. 6.0787 in. 



100 ELEMENTARY APPLIED MATHEMATICS §2 

(16) Referring to Question 7, suppose the speed of the belt 
is 2640 ft. per min. and there is no sKp; (a) how many revolutions 
per minute does the large pulley make? (6) how many does the 
small pulley make? . ( (a) 157.56 r.p.m. 

\(b) 720.30 r.p.m. 



ERRATA 



Study Paper No. 6 

Page 102, Change first two lines of Art. 108, to read as follows: 

"108. If a prism be intersected by a plane parallel to a lateral edge, 

the section so formed by the cutting plane is called a" 
Page 103, Line 20, Change "either base" to read "base of right prism." 
Page 108, Line 16 from bottom, change "area" to "volume." 
Page 109, Line 9, strike out " = ah." 
Page 110, Change answer to Ex. 6 to read "Ans. 13.98, say 14 cu. ft." 

Art. 121, line 3, change "cd" to "cb." 
Page 114, Line 5; change "8070 + " to ".8071 -" line 7, change ".807" 

to ".8071," and "76.61" to 76.6" bottom line, change "r°" to "r\" 
Page 115, Second line of Solution to Ex. 2; change in two places ".807" 

to ".8071," and change "7.^63" to "7.2639;" also, in last line, change 

"7.263" to "7.2639" and change "80.86" to "80.72." 
Page 116, Line 9, change "wiR + r)iR - r)" to "irl(R + r)iR - r);" also, 

4th and 5th lines from bottom, change "201" to "210." 
Page 117, Line 17 from bottom, change "generatrix" to "directrix." 
Page 123, Line 8 of Art 133; read "a"' instead of "a". 
Page 124, Ans. to Ex. 2 should be "593 cu. in." instead of "5930 cu. in.". 
Page 127, Line 17, the second "area" should be "volume." 
Page 129, Line 10 from bottom, change "10^" to "11^." 
Page 130, Ans. to Ex. 5 should be "2368 cu. ft." instead of "339,605 cu. ft." 

Ans. to Ex. 8 should be "23,513 gal." instead of "23, 692 gal." 
Page 133, Last line of Solution, change "86.128" to 86.127." 
Page 136, Line 2, change "5930" to "593;" line 3, change "28497" to 

"2850." Line 9 from bottom, change "5930" to "593." Line 8 from 

bottom, change "15394" to "1540.4." 
Page 145, Ans. to Ex. 7 should be "35.749 cu. yd." instead of "365.88 

cu. yd." 
Page 146, Ans. to Ex. 8 should be "12.179 lb." instead of "12.268 lb." 

Ans. to Ex. 10 should be "1184.9 gal." instead of "319.17 gal." 



ELEMENT AEY APPLIED 
MATHEMATICS 

(PART 3) 



MENSURATION OF SOLIDS 



PRISMS, CYLINDERS, CONES, AND SPHERES 



POLYEDRONS 

104. Every solid has three dimensions — length, breadth, and 
thickness; see Art. 43. A simple example of a solid is shown in 
Fig. 67. Here the sides aa'h'h, hh'c'c, cc'd'd, and aa'd'd are called 
the faces or lateral sides; the end sides abed and a'h'c'd' are called 
the bases. Both ends and all the faces are j^ai, i.e.^ they form 
parts of plane surfaces, and in the illustration, the ends are 
parallel and the opposite faces are also parallel. 




Fig. 67. 

The planes of the faces intersect in right lines to form the 
edges of the soHd; in Fig. 67, the edges formed by the intersec- 
tion of the sides are a' a, b'h, c'c, and d'd. The planes of the faces 
also intersect the planes of the bases in right lines to form the 
edges at the ends of the solid; in Fig. 67, the edges formed by the 

101 



102 ELEMENTARY APPLIED MATHEMATICS 



§2 



intersection of the faces and ends are ab, he, cd, da, a'h' , h'c', 
c'd' , and d'a' . 

105. Prisms. — Any solid whose ends and sides are made up of 
plane surfaces is called a polyedron. If the polyedron has two 
equal and parallel bases, it is called a prism, and if the bases of the 
prism are rectangles or squares, it is called a parallelopiped. 
The solid shown in Fig. 67 is a prism and also a parallelopiped. 
The bases of a prism may be polygons of any shape — regular or 
otherwise — but they must be parallel and equal. 

106. If the planes of the bases are perpendicular to the planes 
of the sides, the lateral edges are perpendicular to the edges of the 
bases, and the prism is called a right prism. Fig. 67 shows a 
right prism, which is also a right parallelopiped. 

107. If a prism be intersected by a plane at right angles to the 
lateral edges, as the plane MN in Fig. 67, the outline of the plane 
figure so formed will be equal to the bases, and the section is 
called a right section or a cross section (usually, the latter). 
In Fig. 67, the plane MN intersects the prism in the cross section 
a^^h^^ c^^ d^^ , which is equal and parallel to the base ahcd. 

108. If a prism be intersected by a plane at right angles, i.e., 
perpendicular to, the bases, the section so formed is called a 

longitudinal section; in Fig. 67, the plane PQ 
intersects the prism in the longitudinal section 
a"h"h"'a"' , the plane FQ being perpendicular 
to both bases. Whatever the shape of the 
prism, every longitudinal section is a rectangle, 
since it can cut only two faces at one time, or 
if it cuts three faces, it must pass through one 
of the edges, thus forming but one side of the 
rectangle. 

109. The altitude of a prism is the perpen- 
dicular distance between the bases. Thus, in 
Fig. 68, suppose that the sides are not at right 
angles to the bases; then In, which represents 

the perpendicular distance between the planes of the bases, is 

the altitude. 

110. Referring to Figs. 67 and 68, it will be observed that the 

sides of any prism are parallelograms, and when the prism is a 

right prism, the sides are rectangles. Since the bases are parallel, 

all the lateral edges are of equal length. 




Fig. 68. 



§2 MENSURATION OF SOLIDS 103 

111. Area and Volume of Prism. — By lateral area is meant the 
area of the outside of the soHd not counting the ends; in the case 
of a prism, it is equal to the area of the faces. 

Referring to Fig. 67, note that if the plane MN is perpendicular 
to one of the faces, it is perpendicular to all of them, and the lines 
of intersection of the plane with the faces are perpendicular to the 
edges; thus, a^^h^^ is perpendicular to a' a and h'h, b^^c^^ is per- 
pendicular to b'b and c'c, etc. Hence, these hues are equal to the 
altitudes of the parallelograms forming the faces, the lengths 
being all equal to the lengths of the lateral edges, which are all 
equal. Consequently, the lateral area of a prism is equivalent to 
that of a rectangle whose length is one of the lateral edges and 
whose altitude is equal to the sum of the lengths of the lines 
formed by the intersection of the plane MN (perpendicular to the 
lateral edges) with the faces of the prism, and this latter is equal 
to the perimeter of the polygon formed by the intersection of 
the plane MN with the faces. 

Let A = the lateral area of the prism, I = length of a lateral 
edge, and p = the perimeter of the polygon formed by a right 
section of the prism = perimeter of either base; then, 

A =pl (1) 

If the area of the ends is also included, the result is called the 
entire area. Let Ae = the entire area and let a = the area of 
one end; then, 

Ae = A -{- 2a = pi + 2a (2) 

112. By volume of a solid is meant the number of cubic units 
that it contains; if the unit of measurement is one inch, then the 
volume is the number of cubic inches that the solid would occupy 
if made of some soft material that could be formed into a cube. 
The phrase cubical contents is frequently used instead of the word 
volume, and the two terms have identically the same meaning. 

Let V = the volume of a prism, a = the area of one of the 
bases, and h = the altitude; then, 

V = ah (1) 

That is, the volume of a prism is equal to the area of the base 
multiplied by the altitude. 

If the base is a rectangle, let I = the length and b = the 
breadth; then, 

V = blh (2) 



104 ELEMENTARY APPLIED MATHEMATICS 



§2 



If the base is a square, let d = the length of one of the sides; 
then, 

V = d% (3) 

If the base is a square and the altitude is equal to one of the 
sides of the base, 

V = h' (4) 

If the prism is a right prism, the altitude is equal to one of the 
lateral edges, and if it is also equal to one of the edges of the base, 
the prism is a cube, and its volume is found by formula (4) . 

Example 1. — A room is 22 feet long, 14 ft. 8 in. wide, and 9 ft. 10 in. 
high; how many cubic feet of air does the room contain? 

Solution. — The cubical contents of the room is, by formula (2), the prod- 
uct of the length, width, and height. 



118 
12 



ft. Hence, F = 22 X ^ X ^^ 



14 ft. 8 in. = 
118 456896 
144 



176 
12 

= 3172 



ft. and 9 ft. 10 

S 

Q cu. ft. Ans. 



Example 2. — A blowpit 20' X 10' X 14' is filled with stock weighing 
65 lb. per cubic foot; what is the weight of the stock? If the stock is 
5.7% fiber, what is the weight of the fiber? 

Solution.— The cubical contents of the blowpit is 20 X 10 X 14 = 2800 
cu. ft. Since 1 cu. ft. of stock weighs 65 lb., the weight of the stock is 2800 
X 65 = 182,000 lb. Ans. 

Since 5.7% of the stock is fiber, the weight of the fiber is 182,000 X .057 
= 10,374 lb. Ans. 

113. Pyramids. — A pyramid is a polyedron having one base, a 
polygon, and whose sides are triangles meeting at a common point 
called the vertex. When a pyramid is 
referred to by letters placed at the vertex 
and at the corners of the base, the letter 
at the vertex is written first, followed by 
a short dash, and then the letters of the 
base. Thus, the pyramid shown in Fig. 
69 would be referred to as the pyramid 
s-abcd. 

Note that the pyramid in Fig. 69 has 
a square base; the sides ash, bsc, csd, and 
dsa are triangles whose vertexes have the 
common point s. The polyedron (pyra- 
mid) in this case is formed by the intersection of five planes — ■ 
four forming the sides and the fifth forming the base. 

If, from the vertex s, a perpendicular be drawn to the base, the 
line so in Fig. 69, this line is the altitude of the pyramid. 




Fig. 69. 



§2 



MENSURATION OF SOLIDS 



105 



If the base of a pyramid is a regular polygon and the projection 
of the vertex upon the base coincides with the geometrical center 
of the base, the pyramid is called a regular pyramid. Suppose 
that the base of the pyramid shown in Fig. 69 is a square and 
that the point o, which is the projection of the vertex s upon the 
base, is the center of the square, then s-abcd is a regular pjrramid, 
and the line so is called the axis of the pyramid. The length of 
so is also equal to altitude of the pyramid. 

If a line sf be drawn from the vertex s perpendicular to one of 
the sides of the base of a regular pyramid, this line is called the 
slant height. There will be as many slant heights as there are 
sides in the polygon forming the base, and for regular pyramids, 
all slant heights are equal; if the pyramid is not regular, some or all 
the slant heights are different. 

114. Area and Volume of Pyramid. — The lateral area of any 

regular pyramid is equal to perimeter of the base multiplied by 

one-half the slant height. This is evident, since the sides are 

sf 
isosceles triangles, and the area of one side is ^ X be, Fig. 69; 

hence, the lateral area 18 ^ X be X n (n = number of sides in 

the base). But n X be = p, the perimeter of the base. Let 

s = the slant height and A. = the lateral area; 

then, 

A = ^X p = hp 

If the pyramid is not regular, calculate the 
area of each face and find the sum. 

115. The volume of any pyramid is equal to 

the area of the base multiplied by one-third the 

altitude. Let h = the altitude = so in Figs. 

69 and 70, V = the volume, and a = the area 

of the base; then, 

h 
V = ^ X a = Iha 

Example. — The sides of the base of a regular triangular pyramid are 
13K in. long and the altitude is 18% in.; what is the entire area and what is 
the cubical contents of the pyramid. 

Solution. — Fig. 70 shows a sketch of the pyramid. Here sf is the slant 
height and so is the altitude. The base abc is an equilateral triangle, and 
of is the apothem, the length of which is (see table, Art. 81) .28868 X 13.25 
= 3.825 in. The triangle sof is a right triangle, right-angled at o, and sf 




106 ELEMENTARY APPLIED MATHEMATICS §2 

= -\/3.8252 + I8.3752 = 18.769 in. The lateral area is, by formula of 

Art. 114, A = 1M69 ^ ^g 25 x 3 = 373.034 sq. in. The area of the base 

13.25^ 
is, by formula (2) of Art. 69, — '-j — VS = 76.021 sq. in. The entire area 

therefore is 373.03 + 76.02 = 449.05 sq. in. Ans. 

The volume is given by the formula above, and is 

18 ^7^ 
V = g X 76.02 = 465.62 cu. in. Ans. 

116. Frustum of a Pyramid. — If a pyramid be intersected by 
a plane parallel to the plane of the base and the top part removed, 
the remaining portion of the pyramid is called a frustum of the 
pyramid. Thus, referring to Fig. 71, the plane a'b'c'd'e' is 

« parallel to the base abcde; now removing that 

,^|'^ part of the pyramid above the intersecting 

// I \\\ plane (here indicated by the dotted lines), the 

„//,_.|-\jf\ remaining part is a frustum of the pyramid 

// H \)X s-ahcde. The two bases of the frustum are 

// ' -J— ^ -^^\ similar plane figures (see Art. 139), and they 

\"l i \ \ ^^ust be parallel. 

/ " \ / ^ good example of a frustum of a pyramid 

/_ V isthebottomof abin: also, a spout. Here the 

be , , 

frustum IS inverted, the large end bemg up, 
and the ends are open. In such cases, the 
frustum has the shape of a solid that would exactly fit the 
opening. 

The altitude of a frustum is the perpendicular distance between 
the bases; it is indicated in Fig. 71 by the dotted line o'o, which 
is a part of the altitude of the pyramid. 

If the frustum is a part of a regular pyramid, the slant height of 
the frustum is that part of the slant height of the pyramid that is 
included between the bases. 

117. In problems relating to frustums, the altitude and the 
length of the sides of both bases are usually given. For a regular 
pyramid, the sides are equal trapezoids, and the lateral area is n 
times the area of one side, where n = the number of sides. Let 
p' = the perimeter of the lower (larger) base and p" = the 
perimeter of the upper base; then, if s = the slant height of the 
frustum of a regular pyramid and A = its lateral area, 

A = Mp' + P") 

118. To find the volume of any frustum of a pyramid, whether 
a regular pyramid or otherwise, let a' = area of lower base, 



§2 MENSURATION OF SOLIDS 107 

a" = area of upper base, h = the altitude, and V = the volume 
of the frustum; then, 

V = ih{a' + a" + Va^X^O 
That is, the volume of a frustum of any pyramid is equal to one-third 
the altitude multiplied by the sum of the areas of the two bases and the 
square root of their product. 

Example. — The lengths of the edges of the lower base of a frustum 
of a triangular pyramid (one having three sides) are 12 in., 15 in., and 18 in.; 
the corresponding edges of the upper base are 7 in., 8% in., and lOK in.; 
if the height of the frustum is 10 inches, what is its volume? 

Solution. — Using the formula of Art. 70, the area of the upper base is 
A =i \/26.25(26.25 - 2 X 7) (26.25 - 2 X 8.75) (26.25 - 2 X 10.5) 
= 30.385 sq. in. since p = 7 + 8.75 + 10.5 = 26.25 in. 

Using the same formula, area of lower base is 89.294 sq. in. 

Then, volume of frustum is 
V = ^ X 10(89.294 + 30.385 + \/89.294 X 30.385) = 572.56 cu. in. Ans. 

119. Prismatoids. — A prismatoid is a polyedron whose ends are 
parallel, but the polygons forming the outline of the ends are not 
equal and may have a different number of 
sides. In Fig. 72 is shown a prismatoid, 
one end of which is a pentagon and the 
other end a quadrilateral. It will be noted 
that four of the sides, abgf, aeif, eihd, and 
cdhg are quadrilaterals, while the side bgc is 
a triangle. These sides are all plane sur- 
faces, and are formed by passing planes 
through the edges of the lower base and the 
vertexes of the upper base. Thus, planes 
passed through the vertex g and the edges 
be and cd intersect to form the lateral edge 
gc; the intersection of these two planes with 
planes passing through the edges ab and de -pia. 72. 

and the vertexes g and / and h and i inter- 
sect in the lateral edges gb and hd; a fifth plane passed through 
ae and the vertexes / and i intersects the last two planes in the 
lateral edges fa and ie. 

If both ends have the same number of sides and are similarly 
situated, the prismatoid is called a prismoid. Thus, the frustum 
of a pyramid is a prismoid, see Fig. 71, since the ends are parallel 
and both contain the same number of sides. Fig. 73 shows 
another prismatoid, the sides ade and 6c/ being triangles; one 
end, abed, is a quadrilateral and the other end is a right line ef. 




108 ELEMENTARY APPLIED MATHEMATICS 



§2 



To find the volume of a prismatoid, let h = the altitude = the 
perpendicular distance between the parallel ends, let' a' = area 
of lower base, a" = area of upper base, and am = area of the 
middle cross section; then the volume of the prismatoid is equal to 
one-sixth the altitude multiplied by the sum of the areas of the upper 
and lower bases and 4 times the area of the 
middle cross section; that is. 

To find the area of the middle section, 
note that a', Fig. 72, is midway between a 
and /, b' is midway between 6 and g, etc. ; 




hence, a'b' 



fg + ab 



b'c' = 



6c + be 



Fig. 73. 



c'd' = 



gh + cd 



etc. 



2 2' 

Knowing the lengths 



of the sides and their position and directions, the area can be 
found. 

This formula is called the prismoidal formula; it may be used 
to calculate the volumes of many solids besides prismatoids. 
Applying it to the example of the last article, the sides of the 

7+12 15 + 8.75 „^_ , 

= 9.5, 5 = 11.875, and 



middle section are 
18 + 10.5 



2 "' 2 

14.25; the area of this triangle is 55.964 sq. in. 

Therefore, the area of the prismoid 
(frustum) is 7 = >^ X 10 (89.294 + 30.385 
+ 4 X 55.964) = 572.56 cu. in.. 

120. The Wedge.— When the base of the 
prismatoid is a rectangle and the end 
parallel to it is a right line parallel to one 
of the edges of the base, the prismatoid is 
called a wedge. Referring to Figs. 73 and 
74, if the base abed is a rectangle and ef 
is parallel to ab, the figure represents a 
wedge. In Fig. 73, the upper base, the 
line ef, is shorter than ab = dc, and the solid is a prismatoid; 
in Fig. 74, ef = ab = dc, and the solid is a triangular prism 
whose bases, eda = fcb, are parallel, and whose altitude is ab. 

The volume of the prismatoid in Fig. 73 may be found by 
applying the prisnjoidal formula; but an easier method is to find 




Fig. 74. 



§2 MENSURATION OF SOLIDS 109 

the sum of parallel sides, divide it by 3, and multiply the quotient 
by the area of a right section — one taken at right angles to the 
parallel sides. Thus, letting a = the area of the right section 
a'e'd', 

y ^ aX {ah -\- dc + ej) ,^. 

o 
When the prismatoid becomes a wedge, ah = dc, and 
y ^ a X (2ah + ej) ^^. 

o 
When a wedge becomes a prism, ah = dc = ef, and 

V = a X ah = ah 

The volume of the wedge may also be calculated by the prismoi- 
dal formula. Thus, let ah = m, be = n, ef = m', and h = the 
altitude; then, for the middle section, the sides parallel to ah or dc 

are equal to ^ ; the sides parallel to he or ad are equal to 

— ^ — = ^; 4 times the area of the middle section is ^ — - 

X K X 4 = mn + m'n; the area of one end is mn, and the area of 
the other end is the line e/ = 0; therefore, by the prismoidal 
formula, ^ = ^ wn + + (mn + m'n) or 

V = lh(2m + m')n. (3) 

For the wedge shown in Fig. 74, m = m', and 

V = ^hmn. (4) 



EXAMPLES 

(1) How many cords of wood are contained in three piles having the fol- 
lowing dimensions: 30' 6" X 7' X 4', 20' X 8' X 4', and 8' X 5' X 4? 
One cord contains 128 cu. ft. Ans. 12f|, say 13 cords. 

(2) A spout has the form of a frustum of a square pyramid ; the bases are 
18 in. and 10 in. square, and the altitude is 28 in. How many square feet 
of sheet iron will be required to make this spout? Ans. 11 sq. ft. 

(3) How many gallons will the spout mentioned in the last example hold 
when filled? Ans. 24.404 gal. 

(4) A trench 20 ft. wide and 36 ft. long is dug; the bottom of the trench 
is level, but the top slopes, me end of the trench being 8 ft. 9 in. deep, the 
other end 5 ft. 3 in. deep, and the slope gradual from end to end. If the 
walls and ends are vertical, how many cubic yards of material were removed? 

Ans. 186f cu. yd. 



110 ELEMENTARY APPLIED MATHEMATICS 



§2 



(5) A freight car is loaded with wood in the following manner, all sticks 
being cut to 4-foot lengths : two piles at each end of the car placed crosswise 
and one pile in the middle (divided into two equal parts) placed lengthwise 
of the car. The average height of the piles at one end is 7' 8", at the other 
end 7' 6", and the height of the pile in the middle is 7'; the inside dimen- 
sions of the car: length, 36 ft.; width, 8 ft. 6 in.; height 8 ft. If an allow- 
ance of 18 in. of the car length be made for spacing between the piles, how 
many cords were placed in the car? Ans. 16.3 cords. 



Suggestion. — The total length of the piles is 36 — 1.5 = 34.5 ft. 
placed crosswise (measured lengthwise of the car) is (34.5 — 4) - 
length of the middle pile is 4 ft. 



The length of each pile 
■ 2 = 153-^ ft.; since the 



(6) A pedestal has a lower base 28" X 40", and upper base 21" X 30", 
and the perpendicular distance between the bases (which are parallel) is 
28 in. How many cubic feet are in the pedestal? Ans. 14.99, say 15 cu. ft. 

(7) How many cubic yards of concrete are required to make a foundation 
wall, the outside dimensions of which are: length, 136 ft.; breadth, 68 ft. 
6 in., the height of the wall being uniformly 7 ft. 9 in. and the thickness 
being 18 in.? Ans. 173.514 cu. yd. 




THE THREE ROUND BODIES 

121. The Cylinder. — If one side ad of a rectangle ahcd, 
Fig. 75, be fixed in position and the rectangle be revolved around 

this fixed side, the opposite side cd will 
generate a curved surface that is called a 
cylinder. The other two sides of the rec- 
tangle will generate the circles hh'h"h"' and 
cc'c"c"', which form the ends or bases, of 
the cylinder. It will be seen that the bases 
are parallel and that they are perpendic- 
ular to the fine ad passing through their 
centers; the line ad is called the axis of the 
cylinder. Any line drawn on the cylinder 
c"' parallel to the axis is called an element 
of the cylindrical surface, or, simply, an 
element; in Fig. 75, he, h'c', etc. are ele- 
ments of the cylinder, and they are evi- 
dently positions occupied by the line he as 
it revolves about the axis db while generating the cylinder. 

122. A cylinder generated as just described is called a cylinder 
of revolution, and since the bases are circles and perpendicular 
to the axis, it is also called a right cylinder with circular base. 

The same rules and formulas that were given for finding the 
area and volume of a prism may be used to find the area and vol- 




FiG. 75. 



§2 



MENSURATION OF SOLIDS 



111 



ume of a cylinder, but since all right sections are circles, these 
rules and formulas may be somewhat simplified. Referring to 
Fig, 76, suppose the cylinder is a cylinder of revolution and that 
it hes on a plane surface; the line (element) ab will then be the line 
of contact of plane and the cylinder, and the plane is said to be 
tangent to the cylinder. Now roll the cylinder on the plane surface 
until the element ab again comes in contact with the plane; the 
distance moved through by the cylinder will evidently be equal 
to the circumference of a circle whose diameter is equal to the 
diameter of the base. During this movement, every element of 




■Trd~ 



Fig. 76. 

the cylinder has come into contact with the plane, and the surface 
thus touched by the cylinder, which is equal to the outside surface 
of the cylinder, is the rectangle aa'b'b, called the development of 
the cyHnder. The outside area of a cylinder, corresponding to 
the lateral area in a prism, is called the convex area, and the sur- 
face is called the convex surface. The diameter of a cyHnder is 
the diameter of a right section, and the altitude of a cylinder 
is the perpendicular distance between the bases ; it is equal to the 
length of the axis in a cylinder of revolution, and it is represented 
by h in Fig. 77. Letting d = the diameter of the cylinder, I = the 
length of the axis, it is plain from Fig. 76 that the convex area of a 
cylinder of revolution is 

A = irdl 
If the cylinder is not a right cylinder, but has parallel bases, as 
in Fig. 77, it may be made a right cylinder by passing a plane 



112 ELEMENTARY APPLIED MATHEMATICS 



§2 



through the cyHnder perpendicular to the axis and just touching 
one edge of the base at c; the part ebjc thus cut off may be placed 
on the other end, as indicated by the dotted lines, and the cylinder 
then becomes a right cylinder e'ecd. The length of the axis is 
the same as before, that is, o'o = o"'o"', hence, A = irdl, as 
before. 




Fig. 77. 



123. If one of the bases is perpendicular to the axis and the 
other is not, for instance, if the cylinder has the form e'bcd, 
Fig. 77, the area is equal to the sum of the areas of the right cylin- 
der e'ecd and the wedge-shaped solid cbe. The latter is evidently 
equal to one-half of a right cylinder whose base is the circle 
ec (center ©") and whose axis is equal to the element eb; that is, 
the area of ebc = ^^-2 X ird X eb = ird X oo", since J-^ X e6 

e'b -H dc 



= oo". But, oo" + o"o' 



00 = 



hence, area of 



cylinder e'bcd is A = irdX o"o"' + Trd X oo" = Trd{oo" + o"o"'), 
or 

(e'b -h dc\ 



A=.d (p^) 



If both bases are oblique to the axis, cut the cylinder by planes 
perpendicular to the axis, calculate the areas of both wedge- 
shaped solids and of the right cylinder included between them, 
and then find the sum of the three results. 



§2 



MENSURATION OF SOLIDS 



113 



Example. — The cylindrical part of a plater roll forms a right cylinder 
17 in. in diameter and 36 in. long; what is its convex area? 

Solution. — Using the formula of Art. 122, 

A = TT X_17 X 36 = 1922 66 sq. in. = 13.352 sq. ft. Ans, 

124, If a right cylinder be intersected by a plane oblique to 
the axis, which cuts the base of the cyUnder, the part thus cut 
off, ifgk, Fig. 78, is called a cylindrical ungula, the word ungula 

having reference to an object shaped ^ 

like a horse's hoof. Let fg be the 
line in which the cutting plane inter- 
sects the base. Bisect fg, and draw 
k'k through the point of bisection o', 
and perpendicular to fg; then k'k is 
a diameter of the base. Let o be 
the middle point of k'k, and ok = ok^ 
= r, the radius of the base = radius of 
cylinder. Let o'f = o'g = a, o'k = b, 
and arc gk = 1/2 arc fkg = <t>; then, 
the convex area of the ungula is, when 
h = the altitude ik, 

A^'f[ii-r)^ + a] (1) 

If the line of intersection pass 
through the center of the base, as 
indicated by the dotted outline in Fig. 78, 6 and a are both equal 

to r, and formula (1) then becomes A = ^[^^ ~ ^^r "^ ^1 

from which 

A = 2rh (2) 

If the cutting plane just touches the other edge of the base, as 
in the case of ebc, Fig. 77, b = 2r and a = 0. Substituting these 

27'/ir irT ~] 

values in formula (1), ^ = ^[(2r - r)y + Oj, since is then 

a semicircle and is equal to rr. Reducing this expression, 

A = ^rh = Yi-Kdh (3) 

which is the same value as was obtained in Art. 123. Note that 

- is the angle okg in radians. 

Example. — Suppose the radius of a right cylinder is 9 in. and that the 
cylinder is cut by a plane in such manner that the altitude of the ungula is 
8 in.; if o = 63^ in., what is the convex area of the ungula? 
8 




114 ELEMENTARY APPLIED MATHEMATICS §2 

Solution. — Referring to Fig. 78, 2a = 2 X 6.5 = IS = fg = chord of arc 
fkg; b = o'k = height of arc = h = r — )4\^^r^ — c^ = 2.775, by formula 

(3), Art. 87; — = one-half the length of the arc divided by the radius = ^• 
Since t = - = ■,„ = .21346, which is very near .21, use formula (3) of 

C io 

Art. 94, and- = g" = g ci e i o(if2) = .8070+. Now substituting in 
formula (1), 

A = ^ ^ ^-^ ^ [(2.775 - 9)0.807 + 6.5] = 76.61 sq. in. Ans. 
2.775 

125. The volume of any cylinder whose bases are parallel is 
equal to the area of the base multiplied by the altitude; it is also 
equal to the area of a right section multipHed by the length of 
the axis. If the cyhnder is one of revolution, the base is equal 
to a right section, and both are equal to the area of a circle whose 
diameter is equal to the diameter of the cylinder; Hkewise, the 
altitude is equal to the length of the axis. Let h = the altitude, 
I = the length of the axis, d — the diameter of the cylinder, 
a = the area of the base; then, 

V^ah = ~dH (1) 

4 

For a cylinder of revolution, 

V = Tr% = ^ d% (2) 

4 

When the bases are not perpendicular to the axis, they have 
the form of an ellipse, and their areas may be calculated by formula 
(1), Art. 99. Thus, referring to Fig. 77, the lower base bjck is 
an ellipse, whose major axis (long diameter) is be, and whose minor 
axis (short diameter) is jk. Knowing these two dimensions 
(which call D and d, respectively), a in formula (1), Art. 99, is 

^ and 6 is ^; hence the area of the ellipse (base) is expressed by 
Trab = TT X -^ X ^ = jDd, and the volume of the cylinder is 

V = '^Ddh (3) 

126. The volume of an ungula is given by the following 
formula, in which the letters have the same meaning as in formula 
(1), Art. 124: 

V = ^[a(3r° - a') + 3r( b- r)<t>] (1) 



§2 



MENSURATION OF SOLIDS 



115 



If the line of intersection fg of the cutting plane with the plane 
of the base passes throu'gh the center, as indicated by the dotted 
outline i'k'g'f, Fig. 78, a = h = r, and formula (1) reduces to 

V = |r2/i (2) 

If the cutting plane just touches the edge of the base, as indi- 
cated by ebjck, Fig. 77, a = 0, & = 2r, ^ = wr, and formula 
(1) reduces to 

(3) 



V = '^r^h 



.807, or <^ = .807 X 9 = 7.263. 



That formula (3) is correct is readily seen, since the ungula ebc 
is equal to one-half the cylinder whose base is the circle ec and 
whose altitude is eh = h in formula (1). 

Example 1. — The shaft of a plater roll is 65 in. long and 8 in. in diameter; 
if made of steel, one cubic inch of which weighs .2836 lb., what is the weight 
of the shaft? 

Solution. — The shaft is a cylinder of revolution, and its volume is 

V ^'^dH ^j X 82 X 65 = -.7854 X 64 X 65 = 3267.264 cu. in. The 
weight of the shaft is 3267.264 X .2836 = 926.6- lb. Ans. 

Example 2. — Referring to the example in Art. 124, what is the volume of 
the ungula? 

Solution. — Here r = 9, h = 8, a = 6.5, 6 (found by calculation) 

= 2.775, and - (found by calculation) = 

Substituting these values in formula (1), 

V = 3 ^ ^^^^ [6.5 (3 X 92 - 6.52) + 3 
X 9(2.775 -9) 7.263] =80.86 cu. in. Ans. 

127. If the cylinder has a hole 
through it, it is called a hollow 
cylinder, pipe or tube. A cross 
section of a hollow cylinder is 
shown in Fig. 79. It is assumed 
that the hole is also a cylinder and 
that its axis coincides with the axis 
of the cylinder that contains the 
hole. The volume of the hollow 
cylinder is evidently equal to the difference of the volumes of the 
cylinder and hole. If I = the length of the cylinder, R = radius 
of cylinder, r = radius of hole, 

V = tRH - irrH = tI(R^ - r') = Trl{R -V r){R - r) 




Fig. 79. 



1{D + d){D - d) 



(1) 



116 ELEMENTARY APPLIED MATHEMATICS §2 

Suppose the cross section in Fig. 79 is that of a tube of thickness 
t; then t = R — r = oa — oh. Suppose further that a circle 
be drawn midway between the inner and outer circles as indicated 
by the dotted line. Then the radius oc of this circle is equal to 

Vm = — ^ — ; its length (circumference) is 2Trm = 27r X — k — 

= t{R -\- r); multiplying this by the thickness t and length I, 
the product will be the volume of a fiat plate having the same 
cubical contents as the tube. Or, since t = R — r, 

V = 2Trm tl = t{R -hr) {R - r) 
which is the same as formula (1). Letting 2rm = dm = diameter 
of middle circle, 

V = Tvdjl (2) 

In other words, the cubical contents of a hollow cylinder, pipe, 
or tube is equal to the continued product of tt = 3 . 1416, the mean 
diameter, the thickness, and the length of the cylinder. 

Example 1. — A cylindrical tank made of wrought iron, a cubic inch of 
which weighs .2778 lb., is to hold 1000 gallons; the tank is to stand with 
the axis vertical, and the diameter (inside) is to be the same as the height; 
what is the diameter of the tank? If the thickness of the shell is 3^ in., 
what is its weight? 

Solution. — Since the diameter equals the height, d = Z in formula (1), 
Art. 125; hence, since there are 231 cu. in. in a gallon, 

= M![^ = ^^ = 1000, or d = ^\^^ = 66.503, say 66J^ in. Ans. 
231 924 ' \3.1416 

The inside diameter is 66.5 in., and since the thickness is .5 in., the out- 
side diameter is 66.6 + 2 X .5 = 67.5 in.; then, by formula (1), Art. 127, 

the cubical contents of the shell is F = ^ X 66.5(67.5 + 66.5) X 1 = 6998.7 

cu. in.; this multiplied by .2778, the weight of a cubic inch of wrought iron, 
is 6998.7 X .2778 = 1944.24, say 1944 lb. Ans. 

Example 2. — Assuming that the shell in the last example is made by roll- 
ing a flat sheet, 3^ in. thick, into a cylindrical form, what should be the 
size of the sheet? 

Solution. — The shape of the sheet will be that of a rectangle, one side of 
which is equal to the height of the tank = 66.5 in., and the other side will 
be equal to the circumference of a circle that is midway between the inside 
and outside circles of a right section of the tank. The diameter of this 
circle is evidently equal to the inside diameter plus the thickness, or 66.5 
+ .5 = 67 in., and its circumference is 3.1416 X 67 = 201.49, say 201^ in. 
Therefore, the sheet must be 201 K in. long and 66 J^ in. wide. Ans. 

Note that the mean diameter of the two circles is — '- — ^ ^ = 67 in., 

the same result as was obtained by adding the thickness to the inside 
diameter. 



V = 



§2 



MENSURATION OF SOLIDS 



117 



128. The term cylinder is not confined to solids having cir- 
cular bases or whose right sections are circles; it is applied to 
solids whose bases have any shape whatever, except those classed 
as prisms, the elements of whose sm'faces are all perpendicular 
to a right section. Such solids may all be generated by keeping 
a line in contact with the outline of a plane figure, then moving 
the line so that it touches every point of the plane figure, and at 
the same time, always remains 
parallel to a given right line. 
The outline of the plane figure 
is called the directrix, and the 
moving line is called the gen- 
eratrix. In Fig. 80, ahcdefgh 
is the directrix and a'a is the 
generatri-x. The generatrix 
moves over the directrix, 
always remaining parallel to 
the fixed line AB, and thus 
generates the cylindrical sur- 
face shown in the figure. 
While, strictly speaking, thes 
directrix should always be a 
curve in order to apply the 
term cylinder to the surface 

generated, it may, nevertheless, be applied to those surfaces in 
which the generatrix consists of both straight lines and curves. 

Keeping in mind this definition of a cylindrical surface and call- 
ing the solid that it bounds a cyhnder, the general formulas 
previously given for the area and volume of a cylinder will apply 
in this case also. 

Example. — What volume of stock is displaced by a beater roll 48 in. 
wide and 60 in. in diameter that is immersed to a depth of 28 in. in the stock? 

SoLtTTiON. — The outline of the stock displaced is a cylinder 48 in. long, 
with equal bases having the shape of a segment of a circle; and a right 
section of this cylinder is a segment of a circle, the radius of the circle bemg 
60 -T- 2 = 30 in., and the height of the segment being 28 in. The volume 
displaced is equal to the area of the segment multiplied by the length of 
the cyhnder. 

To find the area of the segment, first calculate the chord, using formula 
(5), Art. 87, and c = 2V(2 X 30 - 28)28 = 59.867 in. 

Since the angle is very large, use formulas (4) and (6) of Art. 94 to find 
the length of the arc. From formula (4), t' = .19740+, and from formula 




Fig: 80. 



118 ELEMENTARY APPLIED MATHEMATICS 



§2 



(5), I = 90.25 in. The area of the sector is 



90.25 X 30 



1353.75 sq. in.; 



, , . , 59.867 X 2 _ _^ 
area of triangle = ^ = 59.86/ 



sq. in.; and area of segment = 



1353.75 — 59.87 = 1293.88 sq. in. Therefore, volume of cyhnder = volume 
of stock displaced = 1293 88 X 48 = 62,106 cu. in. Ans. 

129. Whenever a cylindrical tank is partly filled with a fluid of 
some kind and is so placed in position that its axis is neither 
vertical nor horizontal, an ungula is formed. The upper surface 
of the fluid is always a horizontal plane, and the intersection of 
this plane with the cylindrical surface forms either an ungula or 
what might be called a frustum of an ungula. Thus, referring to 
Fig. 81, which shows a cylindrical tank having one end higher 




Fig. 81. 

than the other and partly filled with some fluid (stock, for ex- 
ample), the upper surface aa'h'h is a plane, level and parallel to 
the horizontal line eg. If the tank were extended until the 
plane of the top surface just touched the bottom of one end, as 
indicated by the dotted lines, the outline of the fluid contents 
would be the ungula f-acb. The part that is actually in the tank 
has, in this case, the outline acbh'c'a' , which may be called a 
frustum of the ungula f-acb, and its volume is evidently equal to 
the difference between the volume of the ungula f-acb and the 
volume of the ungula f-a'c'h'. The point / can be easily found 
when the length of the tank and the depth of the fluid at each 
end are known. Thus, let I = length of tank, m = dc = depth 
at lower end, n = d'c' = depth at upper end, and x = c'f = the 
additional length of tank required ; then, from the similar triangles 
fed and fe'd', cf : e'f = ed : c'd'. But e'f = x and ef = I + x; 
hence, I -]- x: x ^ 7n: n, or 

nl 



X = 



m — n 



§2 MENSURATION OF SOLIDS 119 

Example.— Referring to Fig. 81, suppose that the tank is 42 in. in diam- 
eter, 12 ft. 6 in. long, and that the upper end is 10 in. higher than the lower 
end; if partly filled with stock, so that the distance ed, measured along the 
tank bottom, is 14^ in-, what is the volume of stock in the tank? 

Solution.— Before the point / can be found, it is necessary to calcula,te 
the distance c'd' = n in th e formula ab ove. Since c'c = 12 ft. 6 in. = 150 in. 
and c'g = 10 in., eg = VlSO^ - 10^ = 149.67 in. Draw ck perpendicular 
to df; then dkc is a right triangle. Since dc is perpendicular to cf, angle dck 
= angle c'cg, because ck is perpendicular to eg, which is parallel to dj. Con- 
sequently, triangles dkc and c'gc are similar, and ck : cd = eg : c'c, or ck : 42 
- 14.25 = 149.67 : 150, from which ck = 27.689 in. = distance from stock 
level to horizontal line eg. Draw e'k' perpendicular to df; then, triangle 
d'k'c' is similar to triangle dkc, since the sides are parallel, and k'g = kc = 
27 689 in. But k'c' = 27.689 - 10 = 17.689 in. Then, from the similar 
triangles d'k'c' and c'gc, e'd' : 150 = 17.689 : 149.67, or c'd' = 17.728 in. In 
the formula above, I = 150, m = 42 - 14.25 = 27.75, and n = 17.728; 

hence x = ^^0 X 17-728 ^ 3 j ^ ^j ^^^ ^j ^ 265.34 + 150 = 

hence, x 27.75-17.728 
415.34 in. 

Calculating first the volume of the ungula f-acb, it is evident that the arc 
acb is greater than a semicircle, since ed is less than the radius oe = 42 ^ 2 
= 21 in. Hence, to find the length of the arc acb, find the length of the arc 
aeb and subtract it from the circumferen ce of whic h it is a part. According 
to Art. 87, db^ = ed X dc, or db = Vl4.25 X 27.75 = 19.886 in. = a in 
formula (1), Art. 126. Since the arc is very large, use formula (2) of Art. 

94 to find the length of the arc aeb. Here t = 2 x 19.886 " ■^^^^^> ^"^ 
40 X 21 X .35829(15 -H6 X .35829^) _^, ^^^ .^_ ^^e circumfer- 
* - 75 + 180 X .358292 + 64 X .35829« 
ence of the circle is 42 X 3.1416 = 131.947 in.; therefore, arc aeb = 131.947 

- 52.242 = 79.705 in. = 2<j> in formula (1), Art. 126, or <l> = 39.8525 in., 
say 39.853 in. Using this formula, h = 415.34, b = 27.75, a = 19.886, 
r = 21, <^ = 39.853, and 

V = -^ ^-^^ [19 886 (3 X 212 _ 19.8862) + 3 x 21(27.75 - 21)39.853 

.3X27.75^ 
= 176,577 cu. in. 

Next calculate the volume of the ungula f-a'c'b'. The arc a'c'b' is less 
than a semicircle, because c'd' is less than the radius o'e'. Here d'b' 

1 7 7*?^ I 

= V'r7.728(42 -l7728y = 20.744; t = 2^2077U ^ ■^^^^°' ^""^ '^ ^ 2 
40 X 21 X .4273(15 + 16 X .4273^) ^ ^g 711 in 

- 2 X 75 -M80 X .42732 + 64 X .4273« 

Now using formula (1), Art. 126, h = 265.34, b = 17.728, a = 20.744, 
r = 21, <^ = 29.711, and 

y,/^_ 268. 13 _ [20.744(3 X 21^ - 20.744^) + 3 X 21(17.728 - 21) 
^ ^ ^^"^^^ X 29.711] = 61,831 cu. in. 

Finally, 7 = F' - F" = 176,577 - 61831 = 114,746, say 114,750 cu. in., 
the volume of the stock in the tank. Ans. 



120 ELEMENTARY APPLIED MATHEMATICS 



§2 



The reader will find it excellent practice to work this entire 
example, performing all the operations herein indicated. 

130. The Cone. — If a right triangle be revolved about one of 
its legs as an axis, the hypotenuse will generate a conical surface, 
and the triangle as a whole will generate 
a solid called a cone. Thus, referring to 
Fig. 82, let soa be a right triangle, right- 
angled at 0, and suppose the triangle to 
be revolved about the leg so; then, the 
hypotenuse sa, and the triangle as a 
whole will generate the solid s-aa'a"b. 
The point s is called the vertex of the 
cone. Any right line drawn from s to 
the base aa'a"h is called an element of 
the cone, more properly, an element of 
the conical surface; thus, sa, sa', sa", 
etc. are elements of the cone in Fig. 82. 
The line so is the axis of the cone, and o 
is the center of the base of the cone. 
Any cone generated in this manner is a cone of revolution; and 
because the base is perpendicular to the axis, the cone is also 
called a right cone. 

Let a right cone be laid on a plane, and suppose the line of con- 
tact of the plane and cone to be sa, Fig. 82; now roll the cone on 





Fig. 83. 



the plane, the vertex s remaining stationary. Each element of 
the cone will come into contact with the plane, and since they are 
all of the same length, the surface of contact thus generated wiU 



§2 MENSURATION OF SOLIDS 121 

be the circular sector saa', Fig. 83, the radius sa being an element 
of the cone, and the length of the arc aa' wiU be equal to the cir- 
cumference of the base of the cone, or 2'k X oa, Fig. 82. Since 
the area of a sector is equal to one-half the product of the radius 
and the length of the arc, it follows that the convex area of a right 
cone is equal to one-half the product of the perimeter of base of 
the cone and the length of an element. The length of an element 
is called the slant height of the cone; representing this by s, and 
letting r = the radius of the base, the convex area is 

A = 3'^s X 2Tr = wrs 

131. If a cone be cut by a plane that intersects the element 
directly opposite the element first cut by the plane, the surface 
formed by the intersection of the plane and cone is an ellipse, 
except when the plane is perpendicular to the axis; in this latter 
case, the section is a circle, which is an ellipse having all its di- 
ameters equal. Referring to Fig. 84, the section a'c'h', which is 
formed by the intersection of a plane perpendicular to the axis so 
of the cone, is a circle; the base, which is « 

also perpendicular to the axis, is also a /i\ 

circle. In Fig. 85, the intersecting plane / j \ 

is not perpendicular to the axis so of / i \ 

the cone, but it intersects the element /{-— i— -A 

ah, which is directly (diametrically) op- cit~—'P—Ab' 

posite the element dc, hence, the section / jc \ 

aed is an ellipse, and the base is also / | \ 

an ellipse. / • \ 

The altitude of a cone is the perpen- /,'-'' — 1 ""'^-^X 
dicular distance between the vertex and t' \ \ 

the base. In the case of a right cone, \ ** y 

the altitude is equal to the length of the ^^-~-~-______---'^ 

axis; it equals so in Figs. 82 and 84. In ^ 

the case of an obhque cone, Fig. 85, the ^^^' ^^' 

base is not perpendicular to the axis, and the altitude is there 

indicated by H, the perpendicular distance between the base 

and the vertex. 

The volume of any cone is equal to one-third the product of 
area of the base by the altitude. Let a = area of base and h = the 
altitude; then, 

V = iah (1) 

If the cone is a cone of revolution and d = the diameter of 



122 ELEMENTARY APPLIED MATHEMATICS 



§2 



the base, the area of the base is o = j^d'^', substituting this value 
of a in formula (1), 

V = ^d^h = .2Q18d% (2) 

If the base is an ellipse, as in Fig. 85, let D = he, the long diam- 
eter, and d = gf, the short diameter; then, the area of the base is 

IT 

jDd = a. Substituting this value of a in formula (1), 



V = ^Ddh = .2Q18Ddh 



(3) 



Example. — Making no allowance for thickness, what is the area of a 
piece of thin sheet metal that will just cover the convex surface of a cone 
with a circular base having a diameter of 33 in. and an altitude of 45 in. ? 
If made of wood weighing 48 lb. per cubic foot, what is the weight of the cone? 

Solution. — The area of the sheet metal will be the same as the area of 



the cone; hence, using the formula of Art. 130, A 



,xfx 



V-+(f) 



= 2484.5 sq. in. Ans. Here the radius is 33 -r- 2, and the slant height is 
the hypotenuse of a right triangle, one leg of which is the radius and the 
other leg is the altitude. 

To find the weight, it is first necessary to calculate the volume. Apply- 
ing formula (2), above, V = .2618 X 33^ X 45 = 12829.5 cu. in. Since 

48 
a cubic foot weighs 48 lb., the weight is evidently 12829.5 X Ty^ = 356.4, 

say 356 lb. Ans. 

132. A frustum of a cone is that portion of the cone included 

between the base and a plane parallel to the base and intersecting 

a the cone. In Fig. 84, a'ahb' is 

a frustum of the cone s —ah] in 
Fig. 85, ahcd is a frustum of 
the cone s — hgcf. In both 
cases, the two bases are parallel; 
but in Fig. 84, both bases are 
perpendicular to the axis, while 
in Fig. 85, they are obUque to 
the axis. It is apparent that a 
frustum of a right cone is always 
a right frustum, since if one 
base is perpendicular to the axis 
(as it must be in the case of a 

right cone) the other base must also be perpendicular to the axis.- 
If a frustum of a right cone be rolled on a plane surface, the 

area thus generated on the plane, called the development of the 




Fig. 85. 



§2 MENSURATION OF SOLIDS 123 

frustum, will have the outhne caa'c', Fig. 83, which may be 
considered to be the development of the frustum in Fig. 84. 
Here sa corresponds to the element m, Fig. 84, and sc corresponds 
to sa'. The area caa'c' is equal to the sector saa' — sector sec' . 
Letting R = radius of lower base and r = radius of upper base, 
perimeter of lower base = 2TrR = arc aa'; perimeter of upper 
base = 2Trr = arc cc' ; then area of frustum is equal to one-half the 
sum of the circumferences of its bases multiplied by the slant height 
of the frustum. Let s = the slant height = a'a, Fig. 84, then 
A = K(27ri^ + 2rr)s, from which 

A = 7rs{R + r) = |s(Z) + d) 

When D and d are the diameters of the bases. 

133. The volume of any frustum of a cone is equal to one-third the 
altitude multiplied by the sum of the areas of the upper base, the 
lower base, and the square root of the product of the areas of the bases. 

Thus, let a' = the area of the lower base, a" = the area of the 
upper base, and h = the altitude = h in Fig. 85; then, 
V = Hh{a' + a" + VV^) (1) 

If the bases are circles, let D = the diameter of the lower 

TT 

base and d = the diameter of the upper base; then, a = t dD^, 

a" = -rd^, and substituting in formula (1) and reducing, 
V = .2618/i(Z)2 + d^ + Dd) (2) 

If the bases are not perpendicular to the axis, they are ellipses; 
letting D' and d' be the long and short diameters of the lower base 
and D" and d" the long and short diameters of the upper base, 

a' = jD'd', a" = jD"d", and substituting in formula (1) and 

reducing, 

V = .2618h{D'd' + D"d" + VD' D"d'd") (3) 

Example. — Find the weight of the plater roll shown in Fig. 86; all right 
sections are circles; the roll is solid; and it is made of cast iron weighing 450 
pounds per cubic foot. 

Solution. — The parts marked A and E are cylinders having the same 
diameter, 8 in., and are therefore equivalent to a single cylinder having a 
diameter of 8 in. and a length of 9 + 14.5 = 23.5 in. The part marked C 
is a cylinder 17 in. in diameter and 36 in. long; the parts marked B and D 
are frustums of cones whose bases have diameters of 12 in. and 17 in. and an 
altitude of 3 in. The volume of the roll will be the sum of the volumes of 
the parts A, B, C, D, and E. The volume of the cylinders may be calcu- 



124 ELEMENTARY APPLIED MATHEMATICS 



§2 



lated by formula (2), Art. 125, and the volume of the frustums by formula 
(2), above. 

Volume of A and E = .7854 X 8^ X (9 + 14.5) = 1181 cu. in. 

Volume of C = .7854 X 17^ X 36 = 8171 

Volume of 5 and D = .2618 X 3(17^ + 122 + 17 X 12) X 2 = 1001 

Total volume =10,353 cu. in. 
The number of cubic feet is .. ' „ , and the weight of the roll is A„p 

X 450 = 2696 lb. Ans. Since the constant 450 is given to only three 
significant figures, the various results were limited to four significant figures. 




Fig. 86. 



Example 2. — Referring to Fig. 85, suppose the long and short diameters 
of the lower base are 16H in. and 11 in., and the long and short diameters 
of the upper base are 6.6 in. and 4.4 in.; if the altitude is 8 in., what is the 
volume of the frustum? 

Solution .— Applying formula (3 ), above, V = .2618 X 8(16.5 X 11 +6.6 
X 4.4 + \/l6.5 X 11 X 6.6 X 4.4 ) = 5930 cu. in. Ans. 

134. The Sphere. — If a semicircle be revolved about its 
diameter, the surface thus generated is called a spherical surface; 
the solid that is bounded by a spherical surface is called a sphere. 
In Fig. 87(a), suppose the semicircle ras to be revolved about its 
diameter rs; it wiU then generate a spherical surface, and the line 
rs is called the axis of the sphere. The middle point o of the 
axis is the center of the sphere. Any right line drawn from the 
center o and ending in the spherical surface, as the line og, is the 
radius of the sphere, and it is evident that all radii of a sphere are 
equal, since they are equal to the radius of the semicircle that 
generates the sphere. Any line that passes through the center 
and is terminated by the spherical surface, as the hne rs, is a 
diameter of the sphere, and any diameter may be taken as the axis. 
The points where the- axis intersects the spherical surface are 
called the poles; in Fig. 87(a); r and s are the poles. 



§2 



MENSURATION OF SOLIDS 



125 






If a sphere be intersected 
by a plane, the figure thus 
formed will be a circle; and 
if the plane passes through 
the center of the sphere, the 
circle is called a great circle; 
otherwise, it is called a small 
circle. In (a) Fig. 87, rash, 
rcsd, resf, etc. are great cir- 
cles; the circles represented 
by the lines kj and mn are 
small circles. All great cir- 
cles are equal; but a small 
circle may have any value 
between that of a great circle 
and 0. Assuming the earth 
to be a perfect sphere, the 
meridians of longitude are 
all great circles, while all 
the parallels of latitude (ex- 
cept the equator) are small 
circles. 

135. The area of a sphere 
is equal to the area of four 
great circles. Let r = the 
radius of a great circle = 
radius of the sphere, and d 
= 2r = the diameter; then, 

^=47rr2 = 7rd!2 = 3. 1416^2 (1) 

The volume of a sphere 
is equal to its area multi- 
plied hy one-third the radius 
= 4xr2 X Ir, or 

y=|7rr3 = i7rd3 = .5236# (2) 

Example. — A certain manu- 
facturer states that a can of his 
paint will cover 100 square feet; 
how many cans will be required 
to give two coats of paint to a 
sphere 100 in. in diameter? If 



126 ELEMENTARY APPLIED MATHEMATICS §2 

the sphere were hollow and its inside diameter were 100 in. how many- 
gallons would it hold? 

Solution. — Applying formula (1) to find the area, which multiply by 2 
since two coats are to be applied, 

A = 3.1416 X 1002 X 2 = 62,832 sq. in. = 62,832 -h 144 = 436+ sq. ft. 
Therefore, the number of cans of paint required is 436 -H 100 = 4.36, say 
4H cans. Ans. 

To find the number of gallons that the sphere will hold, apply formula 
(2) to find the volume, which divide by 231, the number of cubic inches in a- 
gallon; thus, 

V = .5236 X 1003 = 523,600 cu. in., and 523600 -^ 231 = 2266% gal. Ans. 

136. If a sphere be intersected by two parallel planes, the part 
included between the planes is called a spherical segment. 
Referring to (a), Fig. 87, kj and mn represent parallel planes, and 
kamnbj is a spherical segment. It will be noted that a spherical 
segment is somewhat similar to a frustum of a cone, and has two 
bases, kj and mn. If, however, one plane is tangent to the 
sphere, as pq, the spherical segment has but one base. In (a), 
Fig. 87, rkij and rkmnj are spherical segments of one base. The 
perpendicular distance between the bases is the altitude of the 
spherical segment ; thus, il is the altitude of the spherical segment 
kmnj, ir is the altitude of the spherical segment rkij, etc. 

The convex surface of a spherical segment is called a zone; 
thus, the convex surface of kmnj is a zone of two bases, and the 
convex surface of rkij is a zone of one base. Note that a zone 
means a surface — not a solid; there is no such thing as the volume 
of a zone. 

137. The area of a zone is equal to the circumference of a great 
circle of the sphere of which the zone is a part multiplied by the alti- 
tude of the zone. The altitude of a zone is the same as the altitude 
of the spherical segment. Let r = the radius of the sphere, d 
= the diameter of the sphere, and h = the altitude of the zone ; 
then, 

A = 27rrh = wdh (1) 

The volume of a spherical segment is equal to half the sum of its 
bases multiplied by its altitude plus the volume of a sphere of which 
that altitude is the diameter. Thus, letting the letters represent 
the same quantities as before, and letting ri and di be the radius 
and diameter of one base, and r2 and dz the radius and diameter of 
the other base, 

V = }iwiri^ + r2')h + Hrh' = .52S6h[S{ri^ + r^') + h'] (2) 
Also, V = .ld09h[d{di^ + di^) + 4/i2] (3) 



§2 - MENSURATION OF SOLIDS 127 

If the spherical segment have but one base, as rkij, Fig. 87(a), 
make r2 = in formula (2), and 

V = .5236/i(3ri2 + h^) = .lS09h{Sdi^ + 4:h'') (4) 

138. Referring to (6) and (c), Fig. 87, suppose a sphere to be 
generated by revolving the semicircle racs about its diameter rs; 
then, that part of the sphere that is generated by any sector of the 
semicircle is called a spherical sector. In (6) , Fig. 87, the spher- 
ical sector raob is generated by the sector aor of the semicircle; 
the remainder of the sphere, shoa, is also a spherical sector. In 
(c). Fig. 87, aohdoc is a spherical sector generated by the sector 
aoc. The zone forming the convex surface of the spherical 
sector is called the base of the sector. In (6), Fig. 87, the zone 
abr is the base of the spherical sector raob; and in (c). Fig. 87, the 
zone abdc is the base of the spherical sector aobdoc. 

The volume of a spherical sector is equal to the area of the zone 
that forms its base multiplied by one-third the radius of the sphere. 
Since the area of a zone is 2Trrh, the area of a spherical sector is 
2xr/i X }ir, and 

V = ^r% = 2.0944r2/i 
o 

Example 1. — Fig. 88 shows a tank 28 in. long, 12 in. wide, and filled 
with water to a depth of 16 in. A ball 8}i in. in diameter is partly sub- 
merged in the water, the vertical depth, measured on the axis being 6 in'. 
To what additional height, x, will the water level be raised by the ball? 

Solution. — ^Let x = the additional height of the water due to the ball; 
this height will be the same as though an amount of water equal in volume 
to that displaced by the ball had been added to the water in the tank, and 
this volume is equal to the volume of a spherical segment of one base having 
an altitude of 6 in. the diameter of the sphere being 8.5 in. To find the 
radius of the base, let CBDA be a section of the ball, Fig. 89, taken through 
the center; then, CBDA is a great circle, and CD is the axis of the sphere. 
AB is the water level and is 6 in. from the lowest point D of the ball. AB 
is also the diameter of the base of the spherical segment ADB, and in Fig. 
89, is the chord of the arc ACB. One-half of AB = AE is the radius of the 
base of the spherical segment. Now applying the principle of Art. 87, 
AE^ = r-i2 = CE XED = (8.5 - 6) X 6 = 15. Substituting this value 
of Ti^ in formula (4), Art. 137, 

V = .5236 X 6(3 X 15 + 6=) = 254.47 cu. in. 

which is the volume of that part of the ball submerged in the water. The 
amount of water in the tank is 28 X 12 X 16 = 5376 cu. in. Adding to 
this the volume of the spherical segment, 5376 + 254.47 = 5630.47 cu. in. 
Representing the depth of the water after the ball has been placed in it by 
d, the volume occupied by the water and the submerged part of the ball is 



128 ELEMENTARY APPLIED MATHEMATICS 



§2 



28Xl2Xd = 5630.47; from which, d = 5630.47 ^ (28 X 12) = 16.7574- 
in. Hence, the ball raised the water level 16.7574 - 16 = .7574 in. = x. 

Ans. 

If only the value of x had been desired, all that would be necessary to 

ascertain it, is to divide the volume of the spherical segment by the area of 




Fig. 88. 



the bottom of the tank, since the volume of the prism abcdefgh, Fig. 88, must 
be equal to the volume of the submerged part of the ball, and the area 

of the base of the prism is the same as 
the area of the bottom of the tank. 
Hence, 254.47 -h ( 28 X 12) = .7574 - 
in., as before. 

Example 2. — A ball 15 ft. in diameter 

is to be painted in two colors in such a 

manner that the top and bottom zones 

will be of one color and the middle zone 

of another color, all three zones having 

the same altitude. If the middle zone 

is white and the other two black, which 

Fig. 89. color will require the most paint, and what 

will be the area of the surfaces covered? 

SoLTTTiON. — The altitude of each zone is 15 -i- 3 = 5 ft. By formula (1), 

Art. 137, the area of a zone is irdh; since d and h are the same for all three 

zones, their areas are equal, and 

A = 3.1416 X 15 X 5 = 235.62 sq. ft. 




§2 



MENSURATION OF SOLIDS 



129 



Therefore, the area to be covered with the white paint is 235.62 sq. ft., 
and the area to be covered with the black paint is 235.62 X 2 = 471.24 
sq. ft. Ans. 



EXAMPLES 

(1) Fig. 90 shows an iron casting. The part marked C is 8 in. in diam- 
eter and 1 in. thick; part Z) is 5 in. in diameter; part B is 10 in. square and 
1}4 in. thick; there are four strengthening webs marked A, which may be 
considered as triangular prisms, neglecting the small curve formed where 
they join the middle cylinder. 




Fig. 90. 

Taking the weight of a cubic inch of cast iron as .2604 lb., what is the 
weight of the casting? The height over all is 103^ in., and a hole 1}4 in. 
in diameter passes through the center of the casting. Ans. 95.8 lb. 

(2) A wooden ball 18 in. in diameter has a 9-inch hole through it, the axis 
of the hole coinciding with the axis of the ball; what was the original volume 
of the ball, and what is the volume after the hole has been placed in the ball? 
Note that the part removed when the hole was bored is a cylinder and two 
spherical segments of one base, both segments being equal. 

Ans. 3054- cu. in.; 1983+ cu. in. 

(3) One side of a stock chest 12 ft. high is shaped like a semicircle with a 
radius of 5 ft. 2 in.; the side opposite is a right line 10 ft. 4 in. long and is 



130 ELEMENTARY APPLIED MATHEMATICS 



§2 



joined to the semicircular part by right lines 7 ft. long, the angle between 
the long side and the two short sides being a right angle. If the chest stands 
with its axis vertical, what is its capacity in cubic feet when filled to within 
8 in. from the top? Ans. 1295 cu. ft. 

(4) What is the capacity in United States gallons of a vertical cyUndrical 
stock chest 11 ft. 10 in. high and 14 ft. inside diameter, if it is filled to 
within one foot of the top, no allowance being made for the displacement of 
the agitator? Ans. 12,475 gal. 

(5) If the consistency of the stock is 275 pounds per 1000 U. S. gallons in 
the last example, what is the dry weight of the contents of the stock chest? 

Ans. 3431 lb. 

(6) Find the solid contents in cubic feet of the following logs : 
24 logs 16 ft. long, 18 in. at the large end, 15 in. at small end. 
36 logs 15 ft. long, 17 in. at the large end, 13 in. at small end. 
38 logs 12 ft. long, 15 in. at the large end, 12 in. at small end. 
28 logs 20 ft. long, 18 in. at the large end, 14 in. at small end. 

When calculating the cubical contents of logs, it is customary to take ir = 3, 
because the logs are seldom, if ever, exactly round, and it is not only useless 
to take w = 3.1416, but the results will be inaccurate. 

Ans. 339,605 cu. ft. 

(7) A cylindrical tank 28 ft. long and 68 in. in diameter lies with its axis 
horizontal; how many gallons will it hold when filled to within 7.5 in. of the 
top? Ans. 4969 gal. 




Fig. 91. 



(8) A tank reservoir that has the general shape of a frustum of a right 
cone is 16 ft. in diameter at base, 14 ft. 9 in. in diameter at top, and is 18 ft. 
high; how many gallons will it hold when filled to within 14 in. of the top? 

Ans. 23,392 gal. 

(9) A pipe 3% in. inside diameter discharges water at the rate of 5.62 
ft. per sec; how many gallons will it discharge in one hour? The amount 
discharged is evidently equal to the amount required to fill a pipe of the 
same cross-sectional area and having a length equal to 5.62 X 60 X 60 ft. 

Ans. 967.3 gal. 

(10) Find the volume of a frustum of a cone having the dimensions indi- 
cated in Fig. 91. Ans. 130.72 cu. ft. 



§2 MENSURATION OF SOLIDS 131 

SIMILAR FIGURES 

139. Let ABODE, Fig. 92, be any polygon. From any point 
within the polygon, draw lines, called rays, to the vertexes of the 
polygon; and on these rays, lay off distances OA', OB', OC, etc. 

OA' OB' OC' 
in such manner that the ratios ^yv^ = jy^ = yy^ = etc. For 

OA' 
instance, suppose ^^ =f;then, OA' = f OA, OB' = iOB, OC 

= ^OC, etc. Joining the points A', B', C, etc. by right lines, 
another polygon A'B'C'D'E' is obtained that is said to be 
similar to the polygon ABODE. The angles at the vertexes of 




the two polygons that are similarly placed are equal and the 
corresponding sides are proportional; that is, angle A = angle 

Ar ^ r, IP/. A^'^' B'O' O'D' 

A , angle B = angle B , etc., and . p = np = ~rPfr = ^tc. 

Hence, when two polygons are so related that their corresponding 
angles are equal and their corresponding sides are proportional, 
they are similar. 

The point from which the rays are drawn may be taken any- 
where in plane of the figure — either within or without the polygon; 
in fact, it may be taken outside the plane of the figure, provided 
the planes of the two figures are parallel, as in Fig. 93. Here the 
figure ABODEF is similar to the figure A'B'O'D'E'F', and the 
planes of these figures are parallel. If the vertexes of the polygons 
are connected by lines A' A, B'B, etc., they form the edges of a 
prismoid. If the rays OA, OB, etc. be extended sufficiently far to 
enable another prismoid to be formed and the edges of the second 
prismoid are proportional to the first, then the two solids are 
similar. 



132 ELEMENTARY APPLIED MATHEMATICS 



§2 



In the case of similar plane figures or similar solids, lines that 
are similarly placed are called homologous; thus, in similar poly- 
gons, the corresponding sides are homologous. For instance, in 




Fig. 93. 



Fig. 92, AB and A'B', BC and B'C, etc. are homologous sides; 
AC and A'C, EB and E'B', etc. are homologous lines. In Fig. 
93 if there is another solid similar to the one there shown, as 
the solid a'h'c'd'e'j'abcdef, then the edges A' A and a' a, B'B 
and &'6, etc. are homologous; and lines similarly placed, as B'A 
and Va, A'D and a'd, etc. are homologous lines. 

140. Suppose a plane figure to 
be irregular in shape, as A BCD, 
Fig. 94; if from a point within it, a 
sufficient number of rays be drawn, 
and points on these rays be taken 

OA' _ OB' _ PC _ 
sucn xnati /~\ a — /id — r^c* — eijc., 

then the figure A'B'C'D' will be 
similar to the figure ABCD. Any 
lines similarly placed in the two 

figures will be homologous lines; thus, the fines DA and D'A', 

AC and A'C, etc. are homologous lines. 

141. Given any two similar 'plane figures, their 'perimeters are 

proportional tQ the lengths of an'y tivo homologous lines; hence, if 




Fig. 94. 



§2 MENSURATION OF SOLIDS 133 

the perimeter of one of two similar figures is known and the lengths 
of any two homologous lines are also known or can be measured, 
the perimenter of the other figure can be found by proportion. 
Let p and P be the perimeters and I and L be the lengths of two 
homologous lines; then, 

P :p =L :Z 

From which, P = ^ and p = -j-- 

All regular polygons and circles are similar; two ellipses are 
similar when their long and short diameters are proportional; 
two circular segments are similar when the central angles sub- 
tended by their arcs are equal; etc. 

Example. — A cylindrical tank lies in a horizontal position and is filled 
with water to a depth of 31 in. If the inside diameter of the tank is 40 in. 
and it is known that the wetted perimeter of a tank 6 in. in diameter and 
similarly filled is 12.919 in., what is the wetted perimeter of the given tank? 

Solution. — The wetted perimeter is the circular arc touched by the water. 
Since the two tanks are similarly filled, the depths are homologous lines, the 
diameters are also homologous lines, and the wetted perimeters are similar. 
Letting x = the wetted perimeter sought, x : 12.919 = 40 : 6; whence, x = 
12.919 X 40 ^ gg^28, say 863^ in. Ans. 

142. The areas of any two similar plane figures or solids are 

proportional to the squares of any two homologous lines. Since 

all circles are similar, the areas of any two circles are to each other 

as the squares of their radii, the squares of their diameters, the 

squares of the chords of equal arcs, etc. In Fig. 94, if the area of 

ABCD be represented by A and the area of A'B'C'D' by a, and 

the lengths of the homologous lines AC and A'C are known and 

are represented by L and I, 

A :a=U :V 

LP- V' 

from which, A = a 'K-r^, and a = A X j-2' 

Two cylinders are similar when their altitudes are proportional 
to any two homologous lines and the bases are similar ; two cones 
are similar when the base of one is similar to the base of the 
other and the altitudes are proportional to any two homologous 
lines; two prisms are similar when the bases are similar and the 
altitudes are proportional to any two homologous edges or lines 
of the prisms; etc. Therefore, if A is the area of the surface 
of a solid and a is the area of a similar solid (either the convex 
area or the entire area of the solids) and L and I are the lengths 



134 ELEMENTARY APPLIED MATHEMATICS §2 

of any two homologous lines (the diameter or altitude of a 
cylinder, prism, or cone, or the diameter of either base of a fru- 
stum of a cone, or the length of any edge of a prism is a homo- 
logous line when compared with a similar solid), then 

A:a=L^:l^ 

Example. — Referring to the example of Art. 131, what is the area of a 
piece of sheet metal that will exactly cover a cone having a circular base 
that is 26.4 in. in diameter and whose altitude is 36 in.? 

Solution. — Since the area of a cone having a diameter of base of 33 in. 

and an altitude of 45 in. is 2484.5 sq. in., and since -^ = jp = .8, the cones 
are similar, and letting A = the area of the cone here being con- 
sidered, A : 2484.5 = 36^: 45^, from which A = 2484.5 X^^ = 2484.5 (^j 

= 2484.5 (fl^= 2484.5 X .8^ = 1590.1 - sq. in. Ans. 
If the area had been calculated by the formula of Art. 130, 

^ = ^ X ^— - X -vSe^ + ( — ^) = 1590.1 —, the same result as before. 

143. The volumes of any two similar solids are to each other as the 
cubes of their homologous lines. Let V = the volume of one solid 
and V = the volume of a similar solid; then, if L and I are 
homologous lines, 

V :v =L':l' 

from which, V = v X -j = v ( -yj \ ■ 

144. Any cylinder or prism, any pyramid or cone, any frustum 
of a pyramid or cone, and any sphere or spherical segment may 
be regarded as a prismoid, and its volume may be calculated by 
the prismoidal formula. For instance, in the case of a cone, one 
base of the prismoid is 0; hence, comparing the middle section 
with the base, a homologous line of the base will be twice as 
long as the corresponding homologous line of the middle section, 
and the area of the base will be 2^ = 4 times the area of the 
middle section. Letting a — area of the base, the area of 

the middle section is j; and by the prismoidal formula, V 

= |/i (a + 4 X T + 0) = iO'h, which is the same as formula (1), 

Art. 131. In the case of a sphere, both ends of the prismoid are 
points and their areas are 0; the area of the middle section is 
TT^ and the altitude is 2r. Therefore, by the prismoidal formula, 
7 = 1 X 2r(0 + 4 X 7rr2 + 0) = iirr^, which is the same 
as formula (2) , Art. 135. 



§2 MENSURATION OF SOLIDS 135 

If the cross sections of one prismoid are proportional to similar 
cross sections of another prismoid, but the altitudes of the prismoids 
are not in the same proportion, the volumes of the prismoids are 
proportional to the products of the areas of the cross sections by 
the altitudes. For instance, suppose that the dimensions of the 
frustum of two cones are known, that the cross sections of one 
frustum are proportional to the cross sections similarly placed 
in the other frustum, but that the ratio of the altitudes (or of 
any homologous lines not in the section or in planes parallel to 
the sections) is not the same as the ratio of the cross sections; 
then letting A and a be the areas of any two homologous sections 
and H and h the altitudes, V : v = A X H : a X h. Further, 
letting L and I be any two homologous lines in the sections whose 

A /L\ 2 
areas are A and h,— ^ ly ) • From the foregoing proportion, 

V = V {—] -r- Substituting the value of — 
\al h ^ a 



y-^m 



and 



--(1) 



2^ 

H 



145. Suppose that through any point in any solid, three 
planes are passed perpendicular to one another, as the planes 
AB, CD, and EF, Fig. 95; these planes intersect in the right 
lines mn, pg, and rs. If now measurements are taken in these 
planes parallel to the intersecting lines and these measurements 
are proportional to similar measurements taken in another body, 
and this proportion is the same, no matter where the point may 
be situated, the two solids are similar, and their volumes are pro- 
portional to the products of any set of three homologous lines. 
For instance, suppose the point be located at a corner of the base 
of a rectangular parallelopiped; then, a measurement parallel to 
rs, mn, and pq, will correspond to the length, breadth, and thick- 
ness of the parallelopiped, which may be represented byL, B, and T, 
respectively. Homologous lines of a similar parallelopiped may 
be designated by I, h, and t, and 

V : V =L XB XT : I X b X t 

Example. — Referring to example 2, Art. 133, what is the volume of a 
similar frustum of a cone having an altitude of 13^ in. ? 

Solution. — Since the frustums are similar, the volumes of the two frus- 



136 ELEMENTARY APPLIED MATHEMATICS 



§2 



turns are proportional to the cubes of any two homologous lines, say their 

(1 ^ ^\ 3 

= 28497 cu. in. Ans. 




Fig. 95. 

Example 2. — Suppose that the bases of the frustums in the preceding 
example had been similar, the diameters of the upper bases being 11 in. 
and 14}i in., while the altitudes are 8 in. and 12% in. respectively; what is 

the volume of the latter frustum ? 

Solution. — The ratio of the bases 
(and of any section parallel to the 
bases is 14.25 -;- 11 = 1.3 — ; the ratio 
of the altitudes is 12.375 -^ 8 = 1.5 +; 
hence, the volumes are proportional 
to the bases multiplied by the alti- 
tudes, and V = V (j) ^ ^ = 5930 

/14.25\2 12.375 . 

X i .. -, I X — g — = 15394 cu.m. 

Ans. 

Fig. 96. Example 3. — Referring to the ex- 

ample of Art. 141, suppose the 40-inch 
tank is 14 ft. 5 in. long and the 6-inch tank is 3 ft. 8 in. long; how many 
gallons does each hold when filled to the depth specified? 

Solution. — A cross section of the 40-inch tank is shown in Fig. 96. 
The wetted perimeter is the length of the arc ACB, which was found to be 




MENSURATION OF SOLIDS 



137 



86.127 in. long. The area of the segment ACB is equal to the area of the 
sector AC BO + area of triangle AOB. Using the principle of Art. 87, 
AD = s/q X 31 = 16.703 in. Area of sector = 3>^ X 861.27 X 20 = 86.127 
sq. in. Area of triangle = 16.703 X (20 - 9) = 183.73 sq. in. Area of 
segment = 861.27 + 183.73 = 1045 sq. in. Hence, volume of water 
in 40-inch tank is, since 14 ft. 5 in. = 173 in., 1045 X 173 = 180785 cu. in. 
= 180785 -7- 231 = 782.6+ gal. Ans. 
To find the volume of the water in the 6-inch tank, use the proportion 

/ 6\ 2 4 4 
given above, and 782.6: w = 40^ X 173 :62 X 44, or w = 782.6 ( j- j X jyo 

4.4785— gal. Ans. 



SYMMETRICAL FIGURES 

146. Two figures are said to be symmetrical with respect to 
a line, called the axis of symmetry, when any perpendicular 
to the axis that is limited by the outline of the figure is bisected 
by the axis. Thus, referring to Fig. 97, if the perpendiculars 
A' A, B'B, CD, etc. are bisected by the line mn, mn is an axis of 
symmetry, and the two figures 
ABODE A and A'B'C'D'E'A' are 
symmetrical. Now suppose that 
FBCDG and FB'C'D'G are sym- 
metrical with respect to the axis mn, 
the sides FG coinciding; then the 
figure FBCDGD'C'B'F is said to have 
an axis of symmetry. 

If two figures are symmetrical, they 
are evidently equal; thus, in Fig. 97, 
ABODE - A'B'O'D'E'. Conse- 
quently, if any figure have an axis of 
symmetry, this axis bisects the figure ; 
hence, FBODG = FB'O'D'G' = one- 
half FBODGD'O'B'F. It is also 
evident that if the plane of the paper be folded on the line mn, 
A will fall on A', B will fall on B', will fall on 0', etc., and 
ABODE will be superposed on and will coincide with A'B^ 
O'D'E'. Also, the left half of FBODGD'O'B'F will be superposed 
on and will coincide with the right half; therefore, if a 'plane 
figure can he so folded that every point on one side of the line of 
folding will coincide with a point on the other side, the line of folding 
will he an axis of symmetry and will divide the figure into two equal 
parts. 



B 


A 


/ 


F 

\ 


A> 


/ 






\ 


C 












\J 


^ 


6 


E' / 


D 






D' 



B' 



C 



Fig. 97. 



138 ELEMENTARY APPLIED MATHEMATICS 



§2 



147. If the figure have two axes of symmetry, as mn and pq, Fig. 
98, their point of intersection is called the center of symmetry, 
and the center of symmetry is the geometrical center of the fi.gure. 
Every regular polygon has a center of symmetry, which may be 





Fig. 98. 

found by drawing lines from the vertexes to the opposite vertexes, 
when the polygons have an even number of sides, as 4, 6, 8, etc. ; 
but, if the polygons have an odd number of sides, the axes of 
symmetry are found by drawing lines from the vertexes perpen- 
dicular to the sides opposite them. If the number of sides is even. 




Fig. 99. 

a line drawn perpendicular to any side at its middle point will 
bisect the opposite side and be an axis of symmetry. Conse- 
quently, a regular polygon has as many axes of symmetry as it has 
sides. See (J.) and (5), Fig. 99. The figure may be folded on any 
one of these axes and one half will coincide with the other half. 



§2 MENSURATION OF SOLIDS 139 

An isosceles triangle and an arc, sector, or segment of a circle 
has but one axis of symmetry; see {D), Fig. 99. The figure 
whose outline is shown at (C) also has but one axis of symmetry. 
An eUipse has two axes of symmetry, the long and short diameters. 

148. A solid has a plane of symmetry when sections equally 
distant from the plane of symmetry and parallel to it are equal 
and every point in one section has its symmetrical point in the 
other section. A frustrum of any cone has at least one plane of 
symmetry, which includes the axis of the cone and the long 
diameter of either base. If the frustum is that of a right cone, 
the bases are perpendicular to the axis, and there are any num- 
ber of planes of symmetry perpendicular to the bases. 

If a solid has two planes of symmetry, they intersect in a line of 
symmetry, and if it has three planes of symmetry, one of which is 
at right angles to the other two, the three planes intersect in a 
point of symmetry, which is the center of the solid. Thus, in 
Fig. 95, if the plane AB is a plane of symmetry, that part of the 
rectangular parallelepiped in front of the plane is symmetrical to 
that part behind it; if the plane EF is also a plane of symmetry, 
that part of the sohd to the right of the plane is symmetrical to 
that part to the left, and the two planes intersect in the line of 
symmetry pq. If a third plane AB, perpendicular to the other 
two is also a plane of symmetry, that part of the solid below the 
plane is symmetrical to that part above it, and the three planes 
intersect in the point 0, which is the center of the solid. 

149. Now observe that when a body has one plane of sym- 
metry, the plane divides the body into two equal parts, and the 
center of the body lies somewhere in this plane. When the body 
has two planes of symmetry, the center lies in both planes and, 
hence, lies somewhere in the line of their intersection; the two 
planes divide the body into four parts, and if they are perpen- 
dicular to each other, they divide the body into four equal parts. 
When the body has three planes of symmetry, the center lies in all 
three planes, which have only one common point^the point of 
intersection; the three planes divide the body into eight equal 
parts, and if the planes are perpendicular to one another, the 
eight parts are all equal. 

To find the center of a line, bisect the Une and draw a right Hne 
perpendicular to the given line at the point of bisection; this line 
will be an axis of symmetry, provided the line is symmetrical 



140 ELEMENTARY APPLIED MATHEMATICS 



§2 



(as in the case of a right line or circular arc) ; otherwise, it has no 
center, unless it is symmetrical with respect to a point. To be 
syinmetrical with respect to a point, every line drawn through the 
point and limited by the given line or by the perimeter of the 
given figure must be bisected by the point, which is called a 
center of symmetry. Thus, the line shown in Fig. 100 at (a) is 
symmetrical with respect to the center 0, because OA = OA', 





Fig. 100. 

OB = OB', OC = OC, etc., and is the center of the line. For 
the same reason, the parallelogram at (6) is symmetrical with, 
respect to the center 0. The line shown at (c) is not symmetrical 
with respect to a center or to an axis, and therefore has no center. 
Observe that neither the line at (a) nor the parallelogram at (6) 
have an axis of symmetry — neither can be folded on any line so 
one-half can be superposed on the other ; but both have a center of 
symmetry, which is the center of the figure. 



SOLIDS OF REVOLUTION 

150. Center of Gravity. — The center of gravity of a plane 
surface or section is that point at which the surface will balance. 
If the surface have an axis of symmetry, the center of gravity 
(which may be denoted by the abbreviation c. g.) will lie in that 
axis; and if it have two axes of symmetry, the c. g. will be their 



§2 



MENSURATION OF SOLIDS 



141 



point of intersection. The practical way of determining the 
center of gravity of a section is to make a scale drawing of it on 
stiff paper, say Bristol board or cardboard, then carefully cut out 
the section; next suspend the model against a vertical surface 
from a horizontal pin or needle passed through the model; suspend 
the line of a plumb bob from the same pin, and where the line 
crosses the model, draw a line. Now suspend the model from 
some other point and draw a line where the line of the plumb bob 
crosses; the intersection of the two lines will be the center of 




Fig. 101. 



gravity of the plane surface or section. This method of deter- 
mining the c. g. is clearly shown in Fig. 101. The model is first 
drawn and then cut out, including the holes; a pin hole is made at 
A, so that the heavy part of the model will hang lowest; where 
the plumb line crosses, the line mn is drawn. The model is then 
suspended from another point B and another plumb line drawn, 
which crosses mn at 0, and is the c. g. of the section. 

If the section have one axis of symmetry, it is necessary to 
draw but one plumb line, since where this crosses the axis of 
symmetry will be the c. g. It is advisable to have the plumb 
lines cross at as nearly a right angle as possible, since the point of 
intersection will then be easier to determine. 

If the section (model) be placed on a sharp point directly under 



142 ELEMENTARY APPLIED MATHEMATICS 



§2 



the point 0, it will balance; if laid on a knife edge along either of 
the two plumb lines, it will balance. 

151. If any plane section be revolved about a line in that 
section as an axis, the solid so generated is a solid of revolution. 
Some examples of solids of revolution are the right cylinder, right 
cone, and the sphere. The hollow cylinder or tube of Art. 127 
is also a solid of revolution, and may be generated by revolving a 
rectangle about an axis parallel to one of the sides of the rectan- 
gle; see (a). Fig. 102. If a circle be revolved about an axis, the 
resulting solid will be what is called a cylindrical ring or torus; 
see (6), Fig. 102. If the revolving section is a flat ring, as indi- 
cated by the dotted circle in (&), the torus will be hollow. 




Fig. 102. 

Whatever the shape of the revolving section, the volume of the re- 
sulting solid will be equal to the area of the revolving {generating) 
section multiplied by the distance passed through by the center of 
gravity of the section. If r = the perpendicular distance from the 
center of gravity to the axis, and the section make a complete 
revolution, the distance passed through by the center of gravity 
will be the circumference of a circle having a radius r; representing 
this circumference by c, c = 27rr. If a = the area of the gen- 
erating section, the volume of the solid is 

V = 2Trra 

152. Referring to (a). Fig. 102, the center of gravity of the 
section Hes midway between AD and BC, because the rectangle 
has two axes of symmetry, one of which is parallel to and half 
way between AD and BC; hence, if the diameter of the circle 
described by be represented by dm, and AB and AD be repre- 
sented by t and I, respectively, the area of the rectangle is ^ X Z, 
and V = irdmtl, which is the same as formula (2), Art. 127. 



§2 MENSURATION OF SOLIDS 143 

Referring to (6), Fig. 102, the c. g. of the section is the center of 
the generating circle. Let n = the radius of this circle; then the 
area of the generating section is tTi^, and the volume of the torus 
is 

V = 27rViV = 19.7392riV (1) 

If the torus is hollow, let ri = radius of outer circle of section 
and ra = radius of inner circle of section; the area of the section 
is 7r(r2i - rS) = 7r(ri + r^) (r^ - ra), and F = 27rr X 7r(ri + rg) 
(^1 - ^2), or 

V = 19.7392(ri + r2)(ri - r2)r (2) 

Many, if not the majority, of soHds of revolution that occur in 
practice have generating sections that have an axis of symmetry 
perpendicular to the axis of revolution; thus, referring to (c). 
Fig. 102, the sectionA^Ci) has an axis of symmetry O'B perpen- 
dicular to the axis mn about which the section is revolved. The 
e.g. must he somewhere on O'B, and may be located by making a 
model section, on which draw O'B, and then suspending the 
model from a pin as previously described, indicate where the 
plumb hne crosses O'B. Or, if preferred, the model may be 
balanced on a knife edge, and where this edge crosses O'B will 
be the point 0, the e.g. 

Methods of finding by calculation the center of gravity of 
various geometrical figures will be described in the text treating 
on mechanics. 

Example. — What is the volume of a torus, if the diameter over all is 13>^ 
in. and the diameter of a radial section is % in. ? What is its weight if made 
of cast iron, a cubic inch of which weighs .2604 lb. ? 

Solution.— Referring to (6), Fig. 102, O'A = 13.5 -^ 2 = 6.75 in.; 
AB = Ji in.; OA = ]4 -i- 2 = }{e = .4375 in. = ri in formula (1). O'O 
= 6.75 - .4375 = 6.3125 in. = r; then, 

V = 19.7392 X .43752 X 6.3125 = 23.85 - cu. in. Ans. 

The weight = 23.85 X .2604 = 6.211- lb. Ans. 

By radial section is meant a section made by a plane passing 
through the center of the body; such a section includes the axis 
of the body. All the sections shown in Fig. 102 are radial 
sections. 



ELEMENTAEY APPLIED 
MATHEMATICS 

(PART 3) 



EXAMINATION QUESTIONS 

(1) Find (a) the convex area and (6) the entire area of a cone 
of revolution whose base is 22 in. in diameter and whose altitude 
is 38 in. ,^, /(«) 1367.1 sq. in. 

^^^- I (b) 1747.2 sq. in. 

(2) A hopper having somewhat the shape of a frustum of a 
pyramid has the following dimensions: upper end, a rectangle 
44 in. by 28 in.; lower end, a square 8 in. by 8 in.; altitude 36 in. 
How many gallons will it take to fill the hopper? 

Ans. 82.286 gal. 

(3) A piece of cast iron has the shape of a pentagonal prism; 
the ends are regular pentagons, each edge measuring 2j in., and 
the altitude is 4| in. What is its weight, a cubic inch of cast 
iron weighing .2604 lb.? Ans. 10.206 lb. 

(4) The base of a triangular pyramid is an equilateral triangle, 
one edge which measures 5| in. ; the altitude is 9| in. ; what is its 
volume? Ans. 43.116 cu. in. 

(5) A shallow pan has the shape of a frustum of a cone, and 
its inside dimensions are: upper end 14 in. diameter, lower end 
13 in. diameter, and distance between ends 2| in. ; what arc the 
cubical contents of the pan? Ans. 304.31 cu. in. 

(6) The base of a wrought-iron wedge is 10 J in. by 3i in.; the 
upper edge is 7| in. and the altitude is 15 in. Taking the weight 
of a cubic inch of wrought iron as .2778 lb., what is the weight 
of the wedge? Ans. 64.328 lb. 

(7) How many cubic yards are contained in a foundation 
wall whose length is 24 ft. 3 in., breadth is 15 ft. 8 in., inside 
measurements, and thickness is 16 in.? The height of the wall is 
8 ft. 6 in. Ans. 365.88 cu. yd. 

10 145 



146 ELEMENTARY APPLIED MATHEMATICS §2 

(8) A wooden ball has a cylindrical hole extending through it, 
the axis of the hole coinciding with the axis of the ball. If the 
diameter of the ball is 11 in. and of the hole is 4| in., what is 
the weight of the ball, if a cubic inch of the wood weighs .023 lb. ? 

Ans. 12.268 1b. 

(9) A cylindrical tank lies with its axis in a horizontal position; 
it is filled with stock to within 8 in. of the top; its diameter is 
60 in., and its length is 21 ft. 6 in., inside measurements. How 
many gallons of stock are in the tank? Ans. 2907.6 gal. 

(10) A cylindrical tank 48 in. in diameter and 16 ft. long is 
partly filled with stock. The tank lies in such position that 
the level of the liquid just touches the upper end of the dia- 
meter at one end, and cuts the diameter at the other end in its 
middle point. How many gallons of stock are in the tank? 

Ans. 319.17 gal. 

(11) It is desired to make a cylindrical tank to hold 800 gal., 
the height to be 1| times the diameter; what should be the in- 
side diameter and height to the nearest 16th of an inch? 

. ( Diameter, 53^1 in. 

'^''^' I Height, 80 1 in. 

(12) A wrought-iron sheU 9 ft. 7^ in. long is open at both ends; 
it is 54 in. outside diameter and | in. thick. Taking the weight 
of a cubic inch of wrought iron as .2778 lb., what is the weight 
of the shell? Ans. 3362.7 lb. 

(13) If the periphery of an ellipse whose diameters are 37 in. 
and 12| in. is 73.22 in., what is the periphery of an ellipse whose 
diameters are 9i in. and 3| in.? Ans. 18.305 in. 

(14) The area of a certain figure is 430.6 sq. in., and the length 
of a line drawn through the figure is 21 j in.; what is the area of 
a similar figure, if the length of a line similarly placed is ]9| in.? 

Ans. 357.96 sq. in. 

(15) The volume of a frustum of a cone having an altitude of 
8 J in. is 6473 cu. in.; what is the volume of a similar cone having 
an altitude of 6| in.? Ans. 3352 cu. in. 

(16) What is the volume generated by revolving an isosceles 
triangle about an axis parallel to the base, under the following 
conditions? base is 6^ in., the other two sides are each 4| in., 
distance from axis to base is 7 in. The distance from the center 
of gravity of any triangle to the base is one-third the altitude. 

Ans. 513.84 cu. in. 



SECTION 3 

HOW TO EEAD DRAWINGS 



REPRESENTING SOLIDS ON PLANES 



PRELIMINARY EXPLANATIONS 

1. The Picture Plane. — When viewing an object, it is essential 
that the object be illuminated in some manner, either by light 
originating in or on the object itself or by light coming from an- 
other source and being reflected from the object. Light travels in 
right (straight) Hues, called rays, and no matter what their length, 
whether a fraction of an inch or milHons of mUes, every ray of 
light is absolutely straight. The number of rays of light from 
any particular object is infinite, and they extend in every direc- 
tion (in right lines) from the object unless they are stopped 
by some opaque substance that light cannot penetrate. A 
certain number of these rays enter the eye, wherein an image or 
picture of the object is formed and the object is then said to 
be seen. 

If, when viewing an object, a sheet of paper (transparent so 
light can pass through it) be held between the eye and the object 
and the various lines seen on the object are traced on the paper 
with a pen or pencil, the result will be a drawing or picture of the 
object viewed from the position occupied by the eye relative to 
the object. For every change in the position of the eye, there will 
be a change in the shape of the drawing, and for every change 
in distance between the paper and the eye, there will be a 
change in the size of the drawing, these two facts are clearly 
shown in Fig. 1. Here ABCDEFG is a rectangular prismoid- — in 
this case, a frustum of a rectangular pyramid — which contains a 
hole pasing part way through it; 5 is the eye; and P is the 
plane of the paper, called the picture plane. Now imagine 
lines drawn from the points A, B, C, etc. of the object to the 
§3 1 



HOW TO READ DRAWINGS 



§3 



eye; these lines correspoDd to the rays of Hght from the object 
to the eye, and they pierce the picture plane P in the points 
a, b, c, etc. Joining these points by lines as shown, the result is 
the outline abcdefg, which is a drawing or picture of the object 
ABCDEFG. It is evident that if the point S be moved up or 
down the line S'S" or be moved perpendicular to this line in a 
plane perpendicular to the plane of the paper, say in a plane 
parallel to the picture plane, the rays wUl make a different 
angle with the picture plane and the shape of the drawing will 




Fig. 1. 



be different. It is further evident that as the picture plane 
is brought nearer to the eye, the drawing will be smaller, and if 
moved farther fiom the eye, the drawing will be larger, owing 
to the convergence of the rays. 

2. Note that every ray makes an angle with the picture plane 
that is different from that made by any other ray; thus, the angle 
made by the ray AS is different from that made by BS, FS, etc. 
Now assume the picture plane to be so located that it is perpen- 
dicular to the parallel planes of the bases of the frustum and is 
parallel to the parallel edges CB and FG; assume further that the 
point S, called the point of sight, Ues on a ray passing through the 
center of the face CBGF and perpendicular to the picture plane. 
Then, if the picture plane be very near the object and the point of 
sight be a great distance from the object, all the rays will make 



§3 



REPRESENTING SOLIDS ON PLANES 



angles with the picture plane that are very nearly equal to a 
right angle; and, if the point of sight be assumed to be situated 
at an infinite distance from the object, the rays will all be parallel 
and will all be perpendicular to the picture plane. The resulting 
drawing will then be a projection of the object on the picture 
plane, as shown in Fig. 2. Here only one side of the object is 
represented on the drawing, the side CBGF; the entire line AB is 
projected in the single point a, h; the lines DC, EF, and the line 
representing the edge running back from G are also projected in 
single points, since they are all perpendicular to the picture plane. 
It will be noted that the hole does not appear at all. 

3. WhUe the drawing in Fig. 2 shows only one side of the object 
and gives scarcely any intima- 
tion as to the shape of the object, 
it nevertheless possesses several 
advantages over that shown in 
Fig. 1. On the frustum, the lines 
AB and CD, also AD and CB, 
are equal and parallel, but in 
Fig. 1, these lines are neither 
equal nor parallel, and it would 
be very difficult to determine 
from the lengths of the lines ah, 
he, etc. the true lengths of the 
edges AB, BC, etc. In Fig. 2, 
ch and fg are parallel, as they 
should be; ch = CB and fg = 
FG. Further, by using dotted lines to indicate lines that are 
hidden, the depth and diameter of the hole can also be shown. 

4. Projection Drawings.^ — The drawing shown in Fig. 1 is 
called a scenographic projection drawing, a perspective drawing, 
or, simply, a perspective ; the drawing shown in Fig. 2 is called an 
orthographic projection drawing or, simply, a projection drawing. 

In the majority of cases that arise in practice, two views are given 
of the object, one view being taken parallel to the horizontal 
plane and the other parallel to the vertical plane; and when both 
these views are projection drawings and are fully dimensioned, 
they will usually suffice for reproducing the object in its exact 
size. If the object is a complicated machine or machine part, 
three views may be necessary. If, in addition, the object has a 




Fig. 2. 



4 HOW TO READ DRAWINGS §3 

complicated interior arrangement that is hidden by the outside 
surface of the object, interior views (sections) are frequently 
necessary or desirable, in order to obviate the use of too many 
lines, which may make the drawing very hard to read. These 
features will be made clear in what follows. 

Suppose the frustum of Fig. 1 be enclosed in a glass box, as 
illustrated in Fig. 3, and that the front and right-hand sides and 
the top be treated as picture planes, the glass being assumed to be 
transparent. Proj ecting the object on the front plane ilf, the result 
is the outline d"c"J''e", which may be regarded as a drawing of 
the frustum on the picture plane M when the eye is at an infinite 
distance from it. The hnes Dd" , Cc", etc., are the light rays 
from the object to the eye; as applied to drawings, these lines are 
called projectors. In a similar manner, Cc'", Bh'", etc. are 
projectors from the object to the plane side P, and c"'h"'g"'j"' 
is the projection of the frustum on the plane P, considered as a 
picture plane. Likewise, Ee' , Dd' , etc. are projectors from the 
object to the top plane N , and the outline there shown is a pro- 
jection drawing on the plane N, considered as a picture plane. 
Note that the hole is shown in all three views, being represented 
as a circle in the top plane, and as a dotted rectangle in the front 
and side planes. 

The line of intersection made on a plane A by the intersection 
with it of another plane B is called the trace of the plane B; 
Thus, in Fig. 3, the intersection of plane M with plane N is the 
line JK, and JK is the trace of plane M on plane N; likewise, 
JK is the trace of plane N on plane M. Similarly, KL is the 
trace of plane P on plane N, and KI is the trace of plane P on 
plane M. The lines JK, KL, and KI are so important that 
they have received special names : JK is called the front trace ; 
KL is called the side trace; and KI is called the vertical side 
trace. 

If a plane be passed through the line EF (the lower front edge 
of the frustum) parallel to the plane M, the trace of this plane 
on N will be the line e'/'", and the trace on plane P will be the 
line f^f". (A plane is of infinite extent.) It will be observed 
that e'f" is parallel to the projector Ff"; in fact, it is a projec- 
tor from /' to the side trace, which it intersects in /^". Also, 
f^f" is parallel to the projector Ee' and is a projector from/"' 
to the side trace KL, which it intersects in /*" also. Similarly, 
e'' is the projector of e' and e" on the front trace JK, and is the 



§3 



REPRESENTING SOLIDS ON PLANES 



point of intersection of the traces e"e^ and e^/i' of the plane 
h'e''e"H, which has been passed through EH parallel to plane 
P. Consequently, if the position of the projection of any point 
on the planes N and ilf, iV and P, or M and P is known, the 
position of the projection of the point on the third plane can 
easily be found. For example, suppose the positions of c' and c" 
are known, and it is desired to find the position of this point 
on plane P. Draw c'c^" perpendicular to KL, which it intersects 
in c^"; through c'", draw c^^'c'" perpendicular to KL; also 




Fig. 3. 

0"^^ perpendicular to KI, which it intersects in c"^; draw 
c^^c"' perpendicular to KI (and parallel to KL) ; C^'c'" intersects 
c^^c'" in c'", which is the position of the projection of point C 
of the object on the side plane P. If the projections c" and c'" 
had been given and the projection c' on the plane N had been 
desired, project c" on the front trace in c^; project c'" on the 
side trace in c'"", draw C'c' and c'^c' perpendicular respectively 
to JK and KL, and they intersect in c', which is the required 
projection. 

The outlines a'h'c'd'e'f'g'h', d"c^Te",2.ndic"h"'g"'f"^rQ projec- 
tion drawings of the rectangular frustum; they are not suitable as 
they stand for use as working drawings or for any other purpose. 
To make them suitable, imagine the side M of the glass box to 
be hinged to the topside N, and the side P to be hinged to the top 



6 HOW TO READ DRAWINGS §3 

also. Now lift M until its plane coincides with the plane of N, and 
lift P until its plane coincides with the plane of N also ; the result 
will be as shown in Fig. 4. Note that the projectors gr V and 
gi^g'"^ 6'6^^ and ¥"}}'", etc. become a single right line in Fig. 
4; and the same thing is true of the projectors e'e'" and e'^e", 
d'd" and d^d", etc. In other words, g' may be projected 
directly to g'", h' to 6'", etc.; and e' may be projected directly 
to e', d' to d", etc., the only requisite being that the distance between 
g"'j"' and h"'c"' or the distance between e"j" and d"c" must 
be known, and this will be given by measurements made on the 
object. 

Instead of the arrangement of views shown in Fig. 4, another 
arrangement equally correct may be and is frequently employed. 
Referring to Fig. 3, suppose that instead of hinging the side P to 
the top N, the side P is hinged to the front M; then pulling out the 
side P until its plane coincides with the plane of M, lift both 
planes until the common plane of M and P coincides with the 
plane of N. The result will be the arrangement shown in Fig. 5. 
The arrangement shown in Fig. 4 is the more natural, but that 
shown in Fig. 5 may sometimes be more convenient. Either 
Fig. 4 or Fig. 5 is a correct projection drawing of the frustrum, 
and when properly dimensioned, may be used as a working 
drawing. 

5. Names of, and Number of, Views. — Referring to Figs. 4 
and 5, view N is called either a top view or a plan; it is the view 
obtained when the eye is directly over the object. View M is 
called either a front view or a front elevation or, simply, the eleva- 
tion ; it is the view obtained when the eye is directly in front of the 
object. View P is called a side view or a side elevation; it is the 
view obtained when the eye is so situated as to see the object 
from the side, the picture plane being at right angles to the front 
and top planes. 

In both Figs. 4 and 5, three views are given, and, theoretically, 
three views on planes perpendicular to one another are sufficient 
to give the size and shape of any object, since no solid has more 
than three dimensions^ — length, breadth, and thickness. In 
practice, however, it is sometimes advisable to show more than 
three views, because the number of lines on the drawing would 
otherwise be so numerous that the drawing would be very difficult 
to read. Since the glass box in Fig. 3 has six sides, it is possible 
to get six views without modifying the shape of the box. Thus 



§3 



REPRESENTING SOLIDS ON PLANES 



what is termed a bottom view or inverted plan is obtained by 
using TQRI as a picture plane, another side view by using TQSJ 
as a picture plane, and a back view or rear elevation by using 
RQSL as a picture plane. To get these views into proper posi- 
tion, imagine the bottom plane to be hinged to plane M, the side 
planes to be hinged to plane N, and the back plane to be hinged to 
plane N also. Now revolve the bottom plane TQRI downward 



8 








L 




h 












P 






h 


' 




9 






ff 




\ 


V 




»'/ 


] 


9'" 

b'" 


f^^ 














y 




n 






■•* 


'^i 






e 

1 


d 




c 




C" c 


L___^ 


f" 




/ 


1 


\ , 




fl 




V 

C 


f 


/'" 




J 




i" 


c 


r 


K 








M 


ir 
J.. 














J 








II 


I 



















Fig. 4. 

until it coincides with plane M, and then revolve both planes up- 
ward until they coincide with plane iV, likewise, revolve planes 
TQSJ, IRLK, and RQSL upward until they also coincide with 
plane N; the result will be as shown in Fig. 6. Except for the 
letters at the corners, the side views are alike and the front and 
back views are also alike, since the object is symmetrical; the 
bottom view, however, is somewhat different from the top view 
by reason of the dotted lines, which are used because the base of 
the frustum hides everything above it when the frustum is viewed 
from below. 



8 



HOW TO READ DRAWINGS 



§3 



When all six views are shown, the arrangement in Fig. 6 is to be 
preferred. However, the bottom and sides can be imagined to be 
hinged in any other manner desired that will permit all the planes 
to be brought into coincidence with the top plane. Conse- 
quently, the inverted plan may be placed to the right of the 
right side elevation c"'b"'g"'f", to the left of the left side eleva- 
tion d^"a*"A''"e''", or above the rear elevation a"'h'''-g'"%'"\ and the 



M 



o 



d \ I \ c 



IL 



K 




f. 



n 



Fig. 5. 



two side elevations may be placed to the right and left of the 
front elevation, as noted in Fig. 5, if desired. It will be noted 
that the letters at the outside corners on the inverted plan are 
capital letters, corresponding to those at the corners of the base 
of the frustum, because as shown in Fig. 3 the frustum is sup- 
posed to rest on the bottom plane and there are, consequently, 
no projectors from the base to the picture plane. 

6. Working Drawings. — A working drawing is a projection 
drawing on which all necessary dimensions are marked and on 
which all necessary notes are written or printed that are required 



§3 REPRESENTING SOLIDS ON PLANES 9 




[-- 



E 



M 




f 



a b 



Vtll j J/tl^ 



p 



G 



Fig. 6. 



10 



HOW TO READ DRAWINGS 



§3 



in order that the object represented may be made or reproduced. 
A working drawing of the frustum of Fig. 1 is shown in Fig. 7. 
It will be observed that only two views are given; two views are 
all that are required in this case, as the following considerations 
will show: 

The lines mn and jyq are called center lines; in the present case, 
they are axes of symmetry in the plan and Tnn is an axis of sym- 
metry in the front elevation. It is plainly evident that the point 
of intersection of mn and 'pq is the center of the circle that repre- 
sents the plan of the hole. The plan shows that there are two 




^ 


b.a/l 


1 p 


/ \ 


^ 


\ 



4^' pjg^^" -t^^H^ .i~.^^^^^^ 



u. 



-4f- 



^ 



Fig. 7. 



Pig. 8. 



surfaces, ABCD and EFGH, whose projections are rectangles, and 
the elevation shows that these surfaces are planes (j&at surfaces) 
and that they are parallel. If they were not plane surfaces, the 
lines DC and EF in Fig. 7 would not be straight, and if they were 
not parallel, the lines DC and EF would not be parallel. It is 
seen from the dimensions that EF and HG are both 4 in. from the 
center line pq, and that EH and FG are both 2j in. from the 
center line mn; also, AB and DC are both 2| in. from pq, and 
DA and CB are both if^ in. from mn. Since these four dimen- 
sions are given only once, it is inferred that HG, AB, DC, and EF 
are parallel to pq, and that EH, DA, CB, and FG are parallel to 



§3 REPRESENTING SOLIDS ON PLANES 11 

mn, thus making HGFE and ABCD rectangles, since the main 
center lines are always drawn perpendicular to each other. The 
two planes are parallel and are located symmetrically with 
respect to the center lines mn and pq; their distance apart is 
given in the elevation, and is found to be 6^ in. The general 
shape is, consequently, a prismoid, since if the sides were not 
plane surfaces, some of the lines connecting the top and bottom, 
as DE, CF, BG, and AH (in both plan and elevation) would not be 
straight. The hole is IJ in. in diameter and 3 in. deep, and 
extends down from the top. 

To make the object from say, a piece of wood, a rectangular 
prism having a cross section of 4^" X 8" and a length of 6|" 
would be made. On one end, lines would be drawn parallel to 
the long sides and 21 — If^ = /^ in. from them; also, lines 
would be drawn parallel to the short sides and 4 — 2f = If in. 
from them, as indicated in the elevation, Fig. 8. Then the 
material included between the outer edges and the dotted lines 
would be planed off, the result being the frustum. The center of 
the hole could then be located by drawing the diagonals, as indi- 
cated in the plan, or by laying off ah and cd, as indicated, both 
being equal to If^^", and drawing M; then lay off ho or do equal 
to 2f", thus locating the center o. Having found the position of 
o, drill or bore a hole l\" in diameter and 3" deep, the axis of the 
hole being perpendicular to the bases, and the work is completed. 

It will be noticed that no projectors are shown in Figs. 7 or 8; 
they are never shown on finished drawings or on working draw- 
ings. -When reading a drawing, that is, when studying it to find 
out the shape of the object and the details of its parts, the pro- 
jectors can always be imagined to be present; and if the eye does 
not readily perceive the connection between different parts of 
two views, this may usually be found by means of a straightedge 
assisted, perhaps, with dividers to set off distances. This feature 
will be dwelt on more fully later. 



SPECIAL FEATURES PERTAINING TO DRAWINGS 

7. Different Kinds of Lines Used on Drawings. — In Fig. 9, are 
shown five different lines, which are used in the following manner: 

Line I.— This is a full line and may be of any convenient 
weight (by weight is here meant thickness). This line should 
have the same weight wherever used on any particular drawing; 



12 HOW TO READ DRAWINGS §3 

it is employed in all cases when the outline of the object can be 
seen with the eye in the position it is assumed to occupy when 
the view is drawn. This line is used more than any of the others 
and is, consequently, the most important. 

Line II. — This line, called the dotted line, consists of a succes- 
sion of dots or very short dashes; it is used to show the outline of 
parts that cannot be seen by the eye when in its assumed position 
relative to the view being drawn. The inverted plan, or bottom 
view, in Fig. 6 is a good example of the use of the dotted line. 
Viewing the frustum from the bottom all that can be seen is the 
outline of the base, which is drawn with full lines. The top and 
the edges connecting the top and bottom are then represented by 
dotted lines, and likewise the hole. This line is never used for 
any purpose other than to represent invisible outlines. 

I 



Fig. 9. 

Line III. — The broken and dotted line consists of a series of 
long dashes with a single dot or very short dash between the long 
dashes; it is used for center Unes, and is also frequently employed 
to denote the traces of planes, showing where a section has been 
taken. Its use as a center line is shown in Figs. 6 and 7. Its use 
to indicate where a section is taken will be discussed later. 

Line IV. — This broken and dotted liiie consists of a succession 
of long dashes and two dots; it is used for the same purpose as 
line III, and both rarely appear on the same drawing, but when 
they do, line III is used for center lines and line IV to indicate 
where a section has been taken. 

Line V. — The broken line consists of a series of long dashes; it 
is generally used for the same purpose as projectors, to indicate 
the extension or prolongation of lines, but its principal use is for 
dimension lines. It is also sometimes used to indicate the 
extension of lines that are not actually a part of the view being 
drawn. Fig. 7 shows how it is employed in extending lines and 
in dimension lines. 

While the foregoing specifies the manner in which these lines 



§3 REPRESENTING SOLIDS ON PLANES 13 

are most generally used, there are occasional exceptions, particu- 
larly in the case of working drawings. In some drafting offices, 
the regular full line is made very heavy and a light full line 
is used in the same manner as line III; this practice, however, is 
not recommended. Again, it is quite common practice to use 
the light full line for dimension lines, it being broken only where 
the dimension is placed. But, when lines III, IV, and V are 
used, they are usually employed as specified above. 

8. Center Lines. — Center lines are used for two purposes: for 
what may be termed base lines, from which to take measurements; 
and to indicate curved surfaces, particularly circles and cylindrical 
surfaces. They are also used as axes of symmetry. Referring 
to Fig. 6, the center line pq serves two purposes: it is an axis of 
symmetry for the plan and the two side elevations; it shows, 
in connection with the circle in the plan, that the hole is round — 
a right circular cylinder, in fact. The center line mn also serves 
the same two purposes, being an axis of symmetry for the plan, 
the front and rear elevations, and the inverted plan. The point 
of intersection of these two center lines locates the center of the 
circle that represents the top and bottom views of the hole. On 
working drawings, center lines are invariably drawn to define the 
axis of cylindrical surfaces, and their presence on a drawing 
passing through a rectangle is sufficient in itself, as a rule, to show 
that the rectangle is the projection of a cylinder. In the case of a 
circle, two center lines are always drawn at right angles to each 
other, thus locating the center of the circle. The use of center 
lines as base lines is illustrated in Fig. 7; here the dimensions 
2j", 2j", 2f", 2f" etc. which are measured from the center lines 
outward, serve to locate the position of the hole, and they also 
show by their equality that the center lines are axes of symmetry. 

9. Dimension Lines. — Whenever it is possible to avoid it, 
dimension lines are seldom drawn across the face of any view; 
this is to prevent confusion in reading the drawing, and also to 
make the dimensions more prominent. By the use of extension 
lines, the dimension lines and dimensions are very largely kept 
off the different views. Practically the only exception to this 
rule is when giving the diameters of circles and the radii of cir- 
cular arcs. The use of extension lines is shown in Figs. 6 and 7. 
In Fig. 7 it will be noted that the diameter of the hole is printed 
on the front elevation, thus indicating the diameter of the cylin- 



14 HOW TO READ DRAWINGS §3 

drical hole instead of the diameter of the circle in the plan; this is 
done because of the crossing of the center lines on the circle. 

When there is room enough, the dimension lines have arrow- 
heads, one at each end, with the dimension written or printed 
about midway between the ends; but, when the space is limited, 
as in the case of the diameter of the hole in Fig. 7, short arrows 
with their heads pointing toward each other are used, the dimen- 
sion being placed between the arrowheads. In some cases, 
even this will not suffice, the dimension being placed alongside 
the arrow, as illustrated in Fig. 8 in connection with the dimen- 
sion -33- . 

What are termed the general or over all dimensions are those 
relating to the extreme or outer bounding lines of the figure. In 
Fig. 7, the over all dimensions are 8", 4|", and 6j", the first 
two giving the size of the base and the third the height of the 
frustum. The general dimensions include these and also 
5|" and 3t1", which give the size of the top. 

It might be argued that both of the two dimensions, which 
added together make the over all dimension 8" are not necessary, 
since if one is given the other can be found by subtracting from 
8; this is true, but the idea of giving both is to save subtracting, 
which is sometimes awkward. For instance, the distance 
AB, Fig. 7, is Syf"; if the distance from A to the center line 
were given as Ifi", it is not readily apparent that the dis- 
tance from B to the center line is ifi", and this can be found 
only by subtracting ifl" from Syf", a somewhat awkward 
operation and one that takes a certain amount of time, together 
with the possibility of making a mistake. It is for the same reason 
that the over all dimensions are given — ^to obviate the necessity 
of adding the intermediate dimensions. Over all dimensions 
are placed outside of, that is, beyond, all shorter measurements. 

10. Scales. — Up to this point, drawings have been described as 
though every line on them was of the same length as the corre- 
sponding measurements taken on the object. It is obvious that 
this is not feasible in many cases, for not only would it be extremely 
difficult to make the drawing but it would also be very difficult if 
not impossible to read it. The matter of storing and preserving 
drawings must also be considered. Consequently, most of the 
drawings in actual use are made to scale, as it is termed; by 
this is meant that the drawing is smaller than if all the dimensions 
corresponded in length with those of the object. In, such cases, 



§3 REPRESENTING SOLIDS ON PLANES 15 

lines on the drawing are only one-half, one-third, one-fourth, etc. 
as long as those they represent on the object. Sometimes, 
when the object is small and it is desired to bring out certain 
details prominently, the drawing may be made two, three, four, 
etc., times as large as the object. 

Any drawing that has been made accurately, every line being 
carefully measured with a scale, is called a scale drawing. If 
every line on the drawing is of the same length as the line it 
represents on the object, the scale is full size or 12" = 1 ft. If a 
line on the drawing is only half as long as the corresponding line 
on the object, the scale is half size or 6" = 1 ft. if the lines on the 
drawing are only a quarter, an eighth, etc. as long as the corre- 
sponding lines on the object, the scale is quarter size, eighth size 
etc. A quarter-size scale is 3" = 1 ft. because one-fourth of 
12" is 3"; hence, 3" measured on the drawing equal 1 ft. measured 
on the object. If 3" be divided into 12 equal parts, each part 
will represent 1 inch on the object when the drawing is made to a 
scale of 3" = 1 ft. Dividing each of these parts into halves, 
quarters, eighths, etc., these smaller divisions represent halves, 
quarters, eighths, etc. of an inch. 

The scales most commonly used are: full size (12" = 1 ft.), 
half size (6" = 1 ft.), 3" = 1 ft. (quarter size), 1^" = 1 ft. 
(eighth size), and f" = 1 ft. (sixteenth size). The first three 
scales are the most used, and a scale smaller than |" = 1 ft. 
is very seldom needed in drawings of machines. Architects and 
civil engineers use scales very much smaller, a favorite scale 
with architects being 1" = 4 ft. (or |" = 1 ft.). 

Any drawing that has been made to scale should always have 
the scale used marked on it. It sometimes happens that a part 
of the drawing may be made to one scale and a part to another 
scale; in every such case, the scale used should always appear on 
the drawing in connection with the part to which it applies. 

To use a scale to find the actual length of a particular line on 
the object, proceed as follows: suppose the scale is 3" = 1 ft., 
and that an ordinary 12-inch scale, with inches divided into 
halves, quarters, eighths, sixteenths, and thirty-seconds, is used. 
Suppose further that the actual length of the line on the drawing 
that is being measured is 3if in. Since the scale is 3" = 1 ft., 
1 in. on the drawing represents 12 -r- 3 = 4 in. on the object; 
I in. on the drawing represents 1 in. on the object; and sV in- 
on the drawing represents -gV X 4 = |in. on the object. Con- 



16 HOW TO READ DRAWINGS §3 

sequently, 3if in. on the drawing represent 3X4+13X1 = 
12 + If = 13f in. on the object. Had the scale been 1|" = 1 
ft., 1 in. on the drawing = 12 -^ 1| = 8 in. on the object; 3V 
in. on the drawing = tV X 8 = Jin. on the object; and 3|^f in. 
on the drawing represent 3 X 8 + 13 X i = 24 + 3i = 27i in. 
on the object. Observe that since the second scale is twice as 
small as the first, the length of the line when measured to the 
second scale should be twice as long as when measured to the 
first scale; and this is the case, since 13f X 2 = 27|. 

The process of determining the length of a line on the object by 
measuring the corresponding line on the drawing is called scaling 
the drawing, and when a measurement has thus been taken, the 
drawing is said to be scaled. Anj?- drawing may be scaled pro- 
vided the scale to which the drawing was made is known, pro- 
vided further that the drawing has been made accurately and the 
scale used for measuring is accurate. Special rules (scales) are 
made for making drawings to scale and measuring them. 

11. Abbreviations and Notes on Drawings. — When a line is 
drawn from the center to the arc to indicate a radius, the abbre- 
viation r. or rad. is almost invariably written after the dimension, 
thus making it clear that the dimension is the radius of the arc. 
An arrowhead is placed at the end of the dimension line touching 
the arc, but not at the center. The abbreviation D., Dia., 
Diam., d., dia., or diam. is used to follow the dimension indicating 
the diameter of a circle or a cylindrical surface. Thds or ihds 
means threads; thus, 8 thds means 8 threads to the inch, and refers 
to screw threads. The letter/, or fin. uiesms finish, and indicates 
that the surface on which it is written in the drawing is to be 
finished. If only a part of the surface is to be finished, this is 
frequently indicated by drawing a line near to and parallel to the 
line representing the surface, but extending only as far as the sur- 
face is to be finished, and writing on it /. or fin. 

In addition to the dimensions, working drawings usually carry a 
number of "notes," which are written or printed on the drawings 
for the benefit of the workmen. In connection with these notes, 
certain terms are employed that deserve explanation. Thus, the 
word cored means that the hole is to be made by means of a core 
when casting and is not to be finished; hore or hored means that 
the hole is to be cored and finished by boring; drill means that the 
hole is to be made by drilling; ream or reamed means that after the 
hole has been bored or drilled is to be finished by reaming; the 



§3 REPRESENTING SOLIDS ON PLANES 17 

word tap means that the hole is to be threaded, a tap being 
used for this purpose; faced means that the surface is to be 
finished in a lathe or boring mill, or other machine tool that will 
make the surface flat— usually by revolving it against the cutting 
tool; planed means that the surface is to be finished by planing, 
the tool used being a planer, shaper, or a milling machine; grind 
means that the surface is to be finished by grinding; scraped 
means that the surface is to be finished by scraping, that is, 
hand finished with a scraper; tool finish means that after the sur- 
face has been finished on the machine, nothing further is to be 
done to it. Other terms are sometimes used, but they are usually 
self evident and need not be referred to here. 



SECTIONS AND SECTIONAL VIEWS 

12. Why Sections are Used.— While all hidden surfaces and 
parts may be represented on a drawing by dotted lines, it is 
nevertheless frequently advisable to show what are called sec- 
tional views or sections instead of making the drawing in the 
regular manner. The cutting plane may be considered to pass 
through every part of the object that it can touch or its trace may 
be limited to only a short length relating merely to a single 
detail. Sections are usually so taken that lines which would 
appear dotted in a regular projection drawing appear as full lines 
in the sectional view, thus making the drawing easier to read. 
This fact is brought out in Fig. 10, which is a drawing of a clamp- 
ing ring. The center hues mn and pq are axes of symmetry. 
The view in the middle may here be considered as a front eleva- 
tion, the view on the right being a side elevation and that on the 
left a sectional elevation. Note how much clearer the sectional 
elevation is than the side elevation. 

Considering the sectional view, the section is taken along the 
center line mn, the cutting plane being perpendicular to the flat 
surface included between the circles e and g, that is, perpendicular 
to the plane of the paper; that part of the ring to the left of mn is 
then imagined to be removed, and a drawing is made of the re- 
maining part, the eye being situated to the left of mn. Now note 
particularly that whenever a section is taken, any surface touched 
by the cutting plane is indicated on the drawing by cross hatch- 
ing, as it is termed, the cross hatching consisting of parallel hues, 
drawn at an angle, usually 45°, to the horizontal. By using 



18 HOW TO READ DRAWINGS §3 

different kinds of cross hatching, different materials may be in- 
dicated, so that an inspection of the section lines (cross hatch- 
ing) will show at once what material is to be used in making the 
object represented by the drawing. Thus, in Fig. 10, the cross 
hatching there used represents steel, as will be explained pres- 
ently; hence, the clamping ring is to be made of steel. Supposing 
the right-hand view to be omitted, the drawing may then be read 
as follows: 

All dimension lines extending up and down and parallel to the 
center line mn are evidently diameters of circles. The dimension 
9fi" is the diameter of the circle marked a; diameter of circle h 
is 9f|", of circle c, 9if"; of circle d, 9i"; of circle e, 8", and of 
circle/, Q%". Circles h and c are dotted, because, imagining the 
sectional view to be a full view, when the ring is looked at from 
a point to the right of the sectional view, all that part of it to the 
left of CC is hidden. Further, circles h and c are imaginary, 
because the surface they are supposed to indicate is a curved 
surface; still, they are useful in helping to understand the draw- 
ing. The fact that the lines CD and CD' slope shows that the 
ring is a frustum of a cone, or that part of it is which is included 
between DD' and CC That part included between circles e and 
g is flat, and extends downward to form a cylindrical ring, whose 
inside diameter is 5|", outside diameter is 8", and altitude is 
tV + tI = Ig"- Evidently, there is a groove between circles e 
and d; there is also another groove on the under side, as in- 
dicated at E and E', which are at the bottom of another conical 
surface. According to the note, there are 8 holes spaced equally, 
the dotted circle showing that they are all situated at the same 
distance from the center o; and the sectional view shows that 
they pass clear through the ring. The note states that these 
holes are of such size that they may be threaded by tapping 
with a T^th inch tap having 12 threads per inch. The note at 
the bottom of the drawing, which reads "/" all over, means 
that the ring is to be finished all over, that is, no part of it is 
to be left rough. 

13. Standard Sections. — The sections shown in Fig. 11 are 
practically in universal use. The first form, shown at (a), con- 
sists of parallel lines spaced at equal distances apart, all lines 
being of the same weight; this form is always used for cast iron. 
The sectioning for wrought iron is shown at (&) ; it is made by 
drawing light and heavy lines alternately, parallel and equally dis- 



REPRESENTING SOLIDS ON PLANES 



19 



Si M 



P^W" 



^ ^• 



fS 



t--=:D 




pR 



20 



HOW TO READ DRAWINGS 



§3 



tant apart. The sectioning for steel is shown at (c) ; it is made by 
drawing pairs of parallel lines, the distance between two consecu- 
tive pairs being about twice the distance between the lines form- 
ing the pairs. The sectioning for brass is shown at (d); it is 
drawn in the same manner as the sectioning for cast iron, except 
that every other line is broken, being made up of short dashes. 
The sectioning for wood is shown at (e) ; the upper half shows the 



Cast Iron 



Steel 




Brass 




■ 



TFood 



(d) 




form used when the section is taken across the grain, and the 
lower half is used when the section is taken lengthwise, or with the 
grain. 

There are other forms for other materials, but they are not 
much used and there are no other recognized standards, different 
forms being used to indicate the same material in different draft- 
ing rooms. The ones shown in Fig. 11 are well recognized and 
are in nearly universal use. 

14. When the surfaces of two different parts come together in 
a sectional view, the section lines of the two parts are made to 
slope in different directions whenever possible. Thus, referring 
to Fig. 12, which represents a section taken through a bracket, 
which holds up a Une shaft, and its bearing P is the bracket, 
<S is the shaft, C is a collar that is forged to and is a part of 
the shaft, 5 is a bushing (which can be removed and replaced in 
case of wear) , and jB is a collar having the form of a ring, which is 
held in place by a set screw s and keeps the shaft from moving 
lengthwise. As shown by the sectioning, P is made of cast iron, 
B of brass, C and & of steel, and R of wrought iron. Note that 
the section lines for P and B and for R and P slope in different 
directions; it was not possible in this case to have the section 
Hnes for P and B slope in different directions, but since they 
indicate different materials, it does not matter particularly. It 
will also be noticed that the lines at the top and bottom of P 
are ragged, as though P had been broken off at these places; this 



§3 



REPRESENTING SOLIDS ON PLANES 



21 



is exactly what the draftsman intended, since only the details 
of the bearing are to be brought out. For the same reason, 
the shaft is represented as though broken off also; whenever a 




Fig. 12. 



round piece is to be shown as broken off, it is always drawn 
as here indicated. Strictly speaking, the shaft should also have 
been sectioned ; but, by drawing it as here shown and sectioning 
the ends only, the drawing is more readable and is easier to make. 




Fig. 13. 

15. Thin Sections. — When the material to be sectioned is very 
thin, as in the case of sheet metal, boiler plates, etc., or when the 
scale used is so small that the edges of two surfaces in a sectional 
view are very close together, it is then the custom of many drafts- 
men to make the surface to be sectioned entirely black; this is well 



22 



HOW TO READ DRAWINGS 



§3 



illustrated in Fig. 13, which shows the details of a manhole near 
one end of a boiler. Here A is the boiler shell, shown in section by 
the solid black surface, and which has an elliptically shaped hole 
through it, as indicated in the other view and by the lines F; B is 
a steel ring whose inside dimensions are the same as those of the 
hole in the shell, the ring being riveted to the shell in order to 




Section on AB 

Fig. 14. 

strengthen it; C is the manhole cover, also shown in solid black 
sectioning, which is drawn up against the inside of the boiler 
and held in place by the two studs D' and D" and the yokes E' 
and E" , as clearly shown in the two views. Note that when two 
black sectioned surfaces come together, they are separated by a 
thin white line. 

16. Conventional Sections. — A conventional section is one 

that is not drawn in strict accordance with the principles of 
projection; the shaft in Fig. 12 and the rivets in Fig. 13 are exam- 



§3 REPRESENTING SOLIDS ON PLANES 23 

pies of conventional sections. Note that only the end of rivet 
h is shown; this shows that rivet b is behind the cutting plane, 
which cuts rivet a. 

Another example of a conventional section is shown in Fig. 
14. At first glance, it would appear as though the section were 
taken on the center line pq, but an examination of the top view 
shows that if it were not for the part CD of the arms, the section 
would appear to be taken on the center line mn. As a matter of 
fact, the sectional view has been drawn as though the part that 
is cross hatched were a section on mn, and the arms are then 
drawn in between the cross-hatched parts. While this violates 
the rules of projection, the resulting drawing is clear, and this 
method of drawing sectional views in cases similar to this is 
almost universal. To show the shape of a cross section of the arm, 
it is drawn on the arm itself, the section being taken on the line 
EF, which may be located at any convenient point between the 
hub and the rim, but must be perpendicular to the center line of 
the arm. Note that the cross section of the arm has the shape of an 
ellipse and that the arm tapers, the larger end being at the hub. 
As the drawing has been made to a very small scale, the shape of 
a section of the rim of the pulley is not clearly shown in the sec- 
tional view; hence, a section is taken at some other place, as AB, 
and drawn to a larger scale. This section must be a radial 
section, that is, the trace of the cutting plane must coincide with 
a radius, and an examination of the section shows that the rim 
is thicker in the middle than at the ends, the difference between 
the two being called the crown of the pulley. Only one half of 
the pulley is shown in the top view, the lower half not being 
necessary because, the pulley being symmetrical, the lower 
half is exactly the same as the upper half. 



READING DRAWINGS 



VISUALIZING THE OBJECT 

17. A drawing serves several different purposes: it serves to 
preserve in permanent form the shape of some object, and to 
make a permanent record of a view or scene; it is useful, perhaps 
even necessary, in explaining or in understanding the details of a 
complicated machine, the relations of the various parts and their 



24 HOW TO READ DRAWINGS §3 

operation. If properly dimensioned, the drawing may be used to 
show the sizes of the different parts for purposes of comparison 
or for the building of the part or of the complete machine. In 
connection with the first purpose mentioned it may be remarked 
that drawings have been found in France on the walls and ceil- 
ings of caves, which were made from 50,000 to 100,000 years ago. 
These drawings are very crude, but are nevertheless intelligible 
and useful, representing as they do animals that were extinct 
before the beginning of history. 

For a drawing to be useful, it must be so made that the object 
or scene depicted or drawn can be visualized ; that is, after look- 
ing at the drawing, the mind must be able to picture or imagine, 
the object represented in the same manner as though the object 
itself were being viewed by the eye. In the case of a perspective 
drawing, there is no difficulty in visualizing, because the draw- 
ing represents the object or scene as it appears to the eye when 
the eye is in some chosen position relative to the object, the position 
chosen being selected by the draftsman. Projection drawings are 
frequently difficult to visualize, but they can always be pictured 
in the mind's eye if the reader has a knowledge of the principles 
heretofore explained and exercises sufficient patience. While 
progress may be slow at first, continued practice will make one 
proficient in reading a drawing. It takes a great deal of practice 
to associate the letters of a word so that the word is recognized 
as soon as the eye sees it; it is the same way with a drawing, and 
speed in reading a drawing comes only with practice. There is 
no reason for discouragement if, at first, the reader finds it diffi- 
cult to visualize an object represented by a drawing; reading 
drawings is an art that can easily be acquired by application and 
with practice. 

18. When reading a drawing, several facts and principles should 
be continually kept in mind; after a little practice, they will re- 
quire no mental effort to remember them. One of the most 
important of these principles is that any plane (flat) figure, no 
matter what its shape, is projected on any cutting plane that 
intersects the surface at right angles as a right line; in other 
words, a plane figure is projected into the trace of its plane as 
a right line. For example, in Fig. 2, the plane of the top of the 
frustum cuts the picture plane in the line cb, which is the trace 
of the plane AC on the picture plane P. The figure ABCD is 
projected into the trace of its plane on plane P as the right 



§3 READING DRAWINGS 25 

line ch, and the circle within the figure ABCD is also projected 
into the right line cb. Again, a plane figure, no matter what its 
shape, is projected as a similar and equal outline on any plane 
parallel to the plane of the figure. For example, in Fig. 3, the 
plane N is parallel to the plane of the top of the frustum, and the 
projection of the top on plane N is a'h'c'd! , which is equal in all 
respects to ABCD; the projection of the circle is also equal to the 
circle (hole) on the top of the frustum. 

A cylindrical surface is projected on a plane parallel to one of 
its elenients as a quadrilateral, and the projection on a plane per- 
pendicular to an element has the shape of a right section of the 
cylindrical surface. If the cylinder is a right cylinder, the pro- 
jection on a plane parallel to an element is a rectangle, and if a 
right section is a circle or an ellipse, the projection on a plane 
perpendicular to an element is a circle or an ellipse also. 

Always remember that every point on the object or within the 
object is projected into some point on every complete view of the ob- 
ject, and a point on any view may be the projection of one or of any 
number of points on the object. Thus in Fig. 3, the point c'" is 
not only the projection of the point C but also of D and of any 
point on the right line joining C and D; in fact c'" is the projec- 
tion of the entire line CD, because CD is perpendicular to the 
plane of projection KR. If the line joining C and D were not 
straight, but were a plane curve, its projection on plane P would 
be a right line, and if it were shaped like a screw thread, its 
projection on plane P would be a circle. 

It is therefore evident that it is not, in general, possible to 
determine from one view what any point or line represents; but 
two views — one preferably at right angles to the other — will 
usually be sufficient, and three views — one being perpendicular to 
the other two — will always be sufficient to determine the shape of 
any object. If there is any doubt as to what any particular point 
on a view represents, draw (or imagine as drawn) a projector 
from the point to the other view; if the projector coincide with a 
line in the other view, the point is the projection of that line, and 
the point is also the projection of the points of intersection of the 
projector with any line that it cuts in the other view. 

19. Bearing the foregoing in mind, refer to Fig. 4. Looking 
first at the plan (top view), it is seen that it consists of two 
rectangles arranged symmetrically, one within the other, whose 
corners are connected by the right lines a%', b'g', etc. There is 



26 HOW TO READ DRAWINGS §3 

nothing, however, in this view that shows whether the surfaces 
outHned by these rectangles are flat or otherwise, whether 
a'h'c'd' is parallel to h'g'f'e' or not, or whether a'h'c'd' is above 
or below h'g'fe'; there is also nothing to show whether the circle 
represents a hole extending from a'h'c'd' downward or whether 
it represents a cylinder extending upward. All these questions 
are answered by referring to the other two views. Considering 
the front elevation, plane M, point c" is the projection of c' or 6', 
point d" is the projection of d' or a', and line d"c" is the projection 
of d'c' or a'h'. Since d"c" is the only line between d" and e" 
extending across the figure, it is the projection of both d'c' and 
a'h' , c" is the projection of the entire line c'h' , and d" is the pro- 
jection of the entire line d'a'. Since both these lines are straight, 
it follows that a'h'c'd' is a flat surface 'perpendicular to the plane M. 
For similar reasons, h'g'j'e' is also a flat surface perpendicular to 
the plane M. The two surfaces are therefore parallel, and their 
distance apart is the perpendicular distance between the lines 
d"c" and e"j". The dotted rectangle shows that the circle in 
the plan represents a round hole, whose depth is the length of 
i"j", plane M, or i"'j"', plane P. The only feature still undeter- 
mined is the character of the lines joining the corners, that is, 
the lines a'h', h'g', etc. Since these lines are straight in all three 
views, they are straight on the object also, and the sides of the 
object are likewise flat. Consequently, the object is either a 
frustum of a pyramid or a rectangular prismoid, and it has a 
hole in the center extending from the smaller base to within 
about one-half the distance between the bases. The actual 
position of the hole cannot be determined by observation only, 
unless the views are fully dimensioned. Without dimensions, 
the position of the hole can be determined only by the use of 
spacing dividers or a scale. 

20. It is a great help, sometimes, to imagine the paper to be, 
folded on the front and side traces JK and KL, Fig. 4, or on the 
front and vertical side traces JK and KI, Fig. 5, so that planes 
M and P will both be perpendicular to plane N, as in the glass 
box. Then a projector from c" will intersect a projector from 
c'" passing through c', thus locating C on the object; in this 
way, points on the object may be located from the three 
views. 

Suppose that the object were not a frustum of a pyramid, but 
that the top view were the same as before, the top being flat 



§3 



READING DRAWINGS 



27 



and sloping so that all the edges have different lengths. A 
perspective of such an object is shown in Fig. 15 at (a). A pro- 
jection drawing giving three views is shown in the same figure. 
Note that the projectors have been omitted, as is the case in all 
practical drawings. To read this drawing, note first the point a", 
the highest point in the elevation; this corresponds to a' in 
the plan, because if it corresponded to d! , a' would be hidden in 
the front view, and a" d" would be a dotted line instead of a 






Fig. 15. 



full line. For the same reason, h" corresponds to h', and a"h" 
is the projection corresponding to a'h' ) similarly, d" c" is the 
projection corresponding to d'c', d"a" corresponds to d'a', and 
c"h" corresponds to c'h' . The quadrilateral a"h"c"d" is the 
projection of the top of the object on the front plane. By 
reasoning in a similar manner, it will be apparent that the 
quadrilateral a"''b"'c"'d"' is the projection of the top of the 
object on the right side plane. Further, since the bounding fines 
of these projections in all three views are right lines, the surface 



28 HOW TO READ DRAWINGS §3 

is flat. The fact that the surface h'g'j'e' is projected in the 
right lines e"j" and f"g"' in the front and side views indicates 
that this surface is also flat, and since both these lines are horizon- 
tal, the surface represented by h'g'j'e' is a plane surface perpen- 
dicular to both side planes and, consequently, parallel to the 
bottom plane. The object, therefore, has the shape of a prismoid 
that has been cut off at the smaller end by a plane making an 
angle with all three planes of projection. By measurement or 
with the spacing dividers, it will be found the distance of h" 
from e"f' and of d" from e"j" are equal; hence, the points D and 
B on the object, see (a) of the figure, are at the same height 
above the plane of the bottom of the object. 



SOME EXAMPLES IN READING DRAWINGS 

21. Guard for Movable Jaw of Vise. — Fig. 16 shows a guard 
for the movable jaw of a vise. The jaw is made of cast iron, 
while the guard is made of steel and forms a bearing surface for 
the head of the screw that moves the jaw. Three views are 
given: a front view, a bottom view (an inverted plan), and a 
sectional side view. The center mn divides the guard into two 
equal parts and is an axis of symmetry. For convenience, the 
same letters of reference have been used on both halves about 
the center line mn. In what follows the front view will be 
designated as {A), the bottom view as {B), and the sectional side 
view as (C). 

The center line mn is supposed to be perpendicular to the bottom 
plane, and is therefore parallel to the front and side picture 
planes. The line/'/' in {B) is the projection of /a, the semicircle 
aaa, and a/; it is also the projection of /c, rr, and c/; hence, the 
front of the guard, as represented by fcrrcfaa a is flat and per- 
pendicular to the bottom picture plane, as also indicated by the 
line r'" f" in (C). The semicircle bbb is projected in the right 
line h'b' in {B) and is connected to the semicircle aaa by a 
circular arc having a radius of 1| in. This arc is also shown 
in (C), and has the same radius. Consequently, the surface 
represented by ab, a'"b'", and a'b' is a curved surface, and since its 
portions on three planes perpendicular to one another are circular 
arcs, it is a surface of revolution; it is not a part of a spherical 
surface, because the radius of a section perpendicular to the axis 



§3 



READING DRAWINGS 



29 



is not the same as for a section that includes the axis, the radius 
of the former being 1 J in. and of the latter 1| in. To make this 
clear, the head of the screw, which bears against this surface, is 
shown in detail at (a) . The thickness of the guard at the point 6 
is indicated by the length of s'b' and s'"b'" in {B) and (C). The 
edge cr is straight and vertical and is intersected by the cylindrical 
surface of which fc is a part; the intersection is denoted by the 
point c in (A), by the dotted line c'd' in {B), and by the dotted 
line c"d" in (C). The points a and c are both projected into the 




same point on ff in {B) and into/" in (C), because they both lie 
on the same surface and are at the same distance from the center 
line mn. On the back of the guard, there is a shoulder on either 
side of the center line, as indicated by the dotted lines ihge; as 
neither shoulder is touched by the cutting plane, the shoulder 
is not sectioned in (C), but its outline is indicated by i"h"g"e"j"l"; 
the line g"j" in (C) and g'j' in {B) being the projection of the line 
of intersection of the flat surface hg and the cylindrical surface 
ge. The dotted line in the hole in (C) is the edge that extends 
backward from r. The Hne e'k' in (5) is the bottom edge 
(inside) of the shoulder; it is projected into the point e" in 



30 HOW TO READ DRAWINGS §3 

(C) ; this fact shows that the bottom ends of the guard are flat, 
the outline of the surface being fa'h's'e'k'V. 

To assist in visuahzing the guard, a perspective of it is shown 
in (6). By comparing the Hnes on the three views with the cor- 
responding hnes on the perspective, the reader should have no 
difficulty in forming a mental picture of this object. As a 
perspective is seldom given, the reader should make every effort 
to visualize from the three views, without the aid of the perspec- 
tive. The three views will show him the shape of the front, the 
shape of the back, the shape of the top and bottom, and the shape 
of the bearing surface; knowing these, it will then be easy 
to visualize the entire object. 

22. The Mercer Cell. — Fig. 17 shows a Mercer cell, which is 
used for the production of caustic soda and chlorine by elec- 
trolysis. A top view of the cell is shown at (a), and a cross 
section on the line AB is shown at (fc). As indicated by the 
sectioning, the cell is an earthenware crock; it is open at the 
top and has five rectangular openings a in the cylindrical part. 
There is also a cover, shown at (c), which fits into the groove 6. 
Attention is called to the funnel-shaped part d, which is a part of 
the crock; its shape will be readily perceived with the aid of the 
top view (a) . The cylindrical part of the crock is covered with a 
layer of asbestos, extending to just above the top of the rectan- 
gular openings; this is held in place by perforated sheet iron, 
which acts as an electrode, and which is kept in position by three 
iron bands /, the latter being made in three pieces, flanged, and 
bolted, as shown at (c) and {d) . Inside the crock are placed four 
carbon electrodes, which are attached to the disk h, as shown at 
(e); this view shows only two electrodes, the other two being 
behind the two that are seen. The disk h rests on the lugs c, 
shown in views (a) and (6) . At (/) is shown the cover, which fits 
into the groove h. The projection j serves not only as a handle 
but also as an outlet through which the lead i, attached to the disk 
fc at its center, passes. The lead i is connected to the generator, 
and it conducts the current from the generator to the carbons g. 
The chlorine gas escapes through the outlet fc, to which is 
attached the pipe I that conducts the gas to the main, see (c) . 

An elevation of the entire arrangement is shown at (c), and a 
battery of six of these cells placed in a sheet-iron tank is shown at 
{g). This drawing is very simple, and requires no further 
explanation. 



§3 



. READING DRAWINGS 



31 




III! 




Fig. 17. 



32 



HOW TO READ DRAWINGS 



§3 







§3 READING DRAWINGS 33 

23. Rotary Drying Furnace. — Fig. 18 shows in a general way 
the manner in which the black liquor in the soda process is evapor- 
ated and the black ash is obtained in the recovery of the soda. 
The furnace a is mounted on wheels so it can be moved away from 
the rotary incinerator h, for cleaning and repairs. The incinerator 
is a horizontal steel cylinder, lined with fire brick, as indicated in 
the sectional view (a), which is a section on the line AB looking 
toward the right, as indicated by the arrows. Steel tires c are 
attached to the incinerator, and these rest on bearing wheels d, 
the wheels supporting the entire weight of the incinerator. The 
incinerator is caused to revolve by means of the gear and pinion 
arrangement e, the pinion being driven by a system of compound 
gears, which, in turn are driven by the engine m, as indicated in 
the sectional view. The thick liquor is stored in the tank/, from 
which it flows through the pipe g and nozzle h into the rear end of 
the incinerator through the opening j. The front end is also 
open; and as the ash is formed, it gradually accumulates, and 
falls out through this opening into a pit or into a car below it. 
The hot gases pass through the incinerator, through the opening 
j, under and through the boiler k (heating the water therein) , and 
through the stack I into the atmosphere. The details of the 
process should now be clear without further explanation. 

24. Worm Washer.^ — Fig. 19 shows a form of apparatus used 
for washing wood pulp. Two views are given, (&) being an end 
view and (a) a longitudinal section taken on the line AB. The 
apparatus consists of two perforated metal cylinders a, which are 
partly enclosed in a wooden box or, tank, j. The cylinders 
have at one end a tire d, which runs between the flanges of the 
bouble-flanged wheels e. At the other end, the cylinder is sup- 
ported by the hollow trunnion c, which turns in the bearing p. 
The cylinder is thus supported by the wheels e at one end and the 
hollow trunnion at the other end. By referring to view (&), 
it will be noted that the box is divided into two compartments at 
the end k, which is funnel-shaped. Note particularly the shape of 
this partition, indicated by the dotted hne abed efg h in (6) , a part 
of it being circular to permit the cylinder to turn in it, and, at the 
same time, follow the outline of the cylinder. The cylinder being 
thin, heavy iron bands o encircle it at intervals, in order to 
strengthen it. In both cylinders is a sheet-metal worm h, which 
is attached to the head of the closed end of the cyhnder that 
carries the hollow trunnion. At the end gf the trunnion, is a 



34 



HOW TO READ DRAWINGS 



§3 




§3 READING DRAWINGS 35 

large sprocket wheel/; these sprocket wheels are connected by an 
endless belt g, so that both cylinders may revolve together. A 
perforated pipe h extends along the entire length of both cylin- 
ders, and is closed at the farther end. The operation of the 
apparatus is as follows : The stock to be washed flows through the 
pipe g and hollow trunnion c into the cylinder, which, as it 
slowly turns, carries it through the cylinder by means of the 
worm. Water flows through the holes in the perforated pipe h 
and enters the cylinder through its perforations, washing the 
stock and falling through the holes into the open space in the 
box below the cylinder, from whence it is discharged through the 
pipes TO. The washed stock from the cylinders falls into the 
compartments k, and is washed into the exit pipe I by water 
discharged from the pipes i. The slanting boards n keep the 
water from splashing out of the box. 

25. Remark. — The object of the last two figures is to illustrate 
the application of the apparatus described; consequently, the 
drawings fail to show many of the details that would be necessary 
to reproduce the apparatus. Drawings of this kind are a feature 
of printed matter — books, technical journals, catalogs, etc. — 
when the main object of the descriptive matter is to explain 
processes, the operation of machines, etc. 

Assuming that the reader has carefully studied all that precedes 
this article and has conscientiously compared all letters of refer- 
ence with the cuts, he ought now to be able to read intelligently 
most of the drawings that he is likely to encounter. Complicated 
ones may give him trouble, not from lack of knowledge, but from 
lack of practice in reading drawings. He will find, at times, 
that a knowledge of the application or purpose of the object 
drawn is necessary to a complete understanding of the drawing, as 
in Fig. 19, for instance. In any case, he ought to be able to read 
any drawing or to understand any illustration given in this 
course. 



SECTION 4 

ELEMENTS OF PHYSICS 

(PART 1) 



MATTER, MOTION, AND FORCE 



MATTER AND ITS PROPERTIES 

1. Definition. — The word matter, as used in its scientific 
sense, means anything that occupies space. Since any finite 
portion of space has length, breadth (or width), and thickness 
(or height or depth), it can be measured; and matter, which 
occupies a certain space, can also be measured, and in the same 
units. As will shortly be shown, matter can also be measured 
in units other than those used to measure space. 

This property of matter, that it occupies space, is called 
extension, and it is a property (or characteristic) common to all 
matter. Any property that is common to everything is called a 
general property of that thing; hence, extension is a general 
property of matter. 

2. General Properties of Matter.— When it is desired to speak 
of a certain amount of matter without specifying what it is, iron, 
water, air, clay, etc., it is called a body. A body may be of any 
size — so small as not to be visible or as large as the earth or larger 
— it simply means a certain amount of matter. 

In addition to extension, matter possesses a number of other 
general properties: impenetrability, divisibility, indestructi- 
bility, inertia, porosity, compressibility, expansibihty, elasticity, 
mobility, and weight. These properties are common to all 
matter and, consequently, to all bodies. A thorough under- 
standing of the meaning of these words will be a great help in 
understanding physics and chemistry. 

3. Three Forms of Matter.— All matter may exist in one of 
three forms or states: the solid, the liquid, or the gaseous state. 

1 



2 ELEMENTS OF PHYSICS §4 

A solid body is one that retains its shape under ordinary condi- 
tions; thus, wood, stone, clay, iron, etc. are solids under ordinary 
conditions. 

A liquid body (or liquid) is one that does not keep its shape 
unless placed in a vessel to keep it from spreading. It is charac- 
teristic of all liquids that they spread out and conform to the shape 
of the vessel that contains them. Further, if the upper surface 
is free, not touched by any solid, it will, under ordinary condi- 
tions, be a flat, level surface — for all practical purposes, a plane 
perpendicular to a diameter of the earth passing through the 
center of the surface. 

A gaseous body (or gas ) is one that completely fills any closed 
vessel that contains it. If the vessel be connected to another 
closed vessel, say by a pipe having a cock, so that communication 
may be opened or closed between the two vessels, and the cock 
be opened, the gas will fill both vessels completely; and any 
particular amount of space in either vessel, say 1 cubic inch, will 
contain exactly the same weight of gas ; but the weight of a cubic 
inch of the gas will not be the same as in the first vessel before 
the cock was opened. 

Most, if not all, substances can exist in all three states; thus, 
water under ordinary conditions is a liquid, when frozen it is a 
solid (ice), and when vaporized it is a gas (steam). The same is 
true of iron, which can be melted (liquefied) and vaporized, and 
of most other substances. Obviously, a substance can exist in 
but one state at a time. 

4. Divisibility. — Divisibility means that any body may be 
divided into two or more bodies. A solid may be divided by cut- 
ting or by macerating or pulverizing it ; a liquid may be divided 
by pouring some of it into another vessel; a gas may be divided 
as described in the last article. Liquids and gases are frequently 
referred to under the general name of fluid bodies or fluids; 
hence, a fluid may be either a liquid or a gas. There are, of 
course, other ways of dividing bodies than those here mentioned. 

5. Molecules and Atoms. — If an ounce of salt be put into a 
pound of water, the salt will apparently disappear; the salt, 
however, is still in the water, as can be proved by various tests, 
as, for instance, by tasting. Here the water is said to dissolve 
the salt. What really occurred was that the water divided the salt 
into particles so small that they were no longer visible. By 



§4 MATTER, MOTION, AND FORCE 3 

stirring the water, the salt particles may be made to occupy 
every part of the water. If more water is added, the salty taste 
becomes less strong, and finally becomes indistinguishable; but 
the salt is still there, though it has been divided into extremely 
small particles. In accordance with modern theory, the division 
cannot be carried on indefinitely, for when a certain stage has been 
reached, any further division will change the nature of the body. 
For example, it is shown in chemistry that salt is a compound 
of sodium and chlorine, one part of each. When the division 
reaches a certain point, the particles can no longer be divided by 
any of the means heretofore mentfoned, and they are then called 
molecules. If a molecule be divided by chemical action or by 
the action of electricity or heat, the parts are called atoms. In 
the case of salt, the molecule when divided becomes two atoms, 
one being sodium and the other chlorine. It has not been found 
possible so to divide an atom, and it is therefore assumed that an 
atom is indivisible. Atoms do not usually exist in a free state, 
two or more being united to form a molecule. 

A molecule, then, may be defined as the smallest particle of 
matter that can exist .without changing the nature of the sub- 
stance of which it forms a part; thus, a molecule of water consists 
of two atoms of hydrogen and one atom of oxygen, and if a 
molecule of water be divided, it will be no longer water, but three 
atoms, which will immediately reunite to form water again or 
unite among themselves to form molecules of hydrogen and 
oxygen. Molecules are exceedingly small, so small that they have 
never been seen and probably never will be seen. They are as- 
sumed to be spherical in shape and are known to be in very rapid 
motion, moving in all directions and constantly colliding. The 
molecules of a solid are more numerous and move through shorter 
distances than those of a liquid, and those of a liquid are more 
numerous and move through shorter distances than those of a 
gas. 

6. Porosity. — All matter is porous, that is, the molecules do not 
completely fill the space occupied by the molecules; otherwise, 
there could be no movement of the molecules. Even though 
they may not be visible to the naked eye or under the highest 
powered microscope, all bodies have minute pores or channels, 
as is proved by the fact that water, for example, can be forced 
through a highly polished, thin steel plate. If matter were not 
porous, it would not be possible to force light or fluids through a 



4 ELEMENTS OF PHYSICS §4 

sheet of metal, no matter how thin it might be; but a piece of 
gold leaf 1 /300,000th of an inch thick will allow light to pass 
through it. 

7. Impenetrability.^ — Impenetrability refers to that property 
whereby two bodies cannot occupy exactly the same space at the 
same time. If a stone be placed in a vessel partly filled with 
water, the level of the water will be raised, showing that the stone 
has taken the place of the same amount of water. Advantage 
may be taken of this fact to find the volume of any irregular 
solid that is not dissovled in water. For instance, knowing the 
volume of the water before and after the solid is immersed in it, 
the difference will be the volume of the solid. If a solid be dis- 
solved in a liquid, the volume may be the same as before, as, for 
example, dissolving a certain amount of sugar in a cup full of tea. 
This does not mean, however, that water is penetrable; the experi- 
ment simply shows that water is porous, and that the molecules of 
sugar have crowded between the molecules of water and occupy 
some of the space not filled by the water molecules. Again, if 
50 c.c. (cubic centimeters) of alcohol are mixed with 50 c.c. of 
water, the volume of the mixture will not be 100 c.c, but 97 
c.c; this shows that one of the hquids is more porous than the 
other, with the result that the combined volume is less than 
the sum of the volumes before mixing. 

8. Compressibility. — Compressibility refers to that property 
whereby bodies can be made to occupy a smaller space. Gases 
are very compressible; liquids and solids are only slightly com- 
pressible, but there is no substance that is incompressible. This 
necessarily follows from the fact that matter is porous, and the 
further fact that the molecules can be brought closer together 
and be made to occupy a smaller space than before. 

9. Expansibility." — Expansibility refers to that property whereby 
the molecules can be forced farther apart and be made to occupy 
a greater space. All gases tend to expand and occupy a greater 
volume; but liquids and solids can usually be made to expand 
only by means of heat, most of them expanding when the 
temperature is raised, although a few expand under certain 
circumstances when the temperature is lowered. 

10. Elasticity. — Elasticity is the name of the property by 
which, after a body has been distorted (shape changed) in any 
way and by any means (except division) the body tends to resume 



§4 MATTER, MOTION, AND FORCE 5 

its former shape when the cause that produced the distortion is 
removed and the former conditions are restored. While the 
body may not entirely regain its former shape, it always has a 
tendency that way and will partly regain it. A piece of rubber, 
for instance, may be stretched, and if not stretched too much, 
will resume its former shape when released. A ball of moist 
clay or putty may be pressed and made to assume and retain any 
shape but it does not retain exactly the shape it had when the 
pressure was removed. All fluids and some solids, such as steel, 
ivory, glass, etc., are very elastic. 

11. Mobility. — The distance between one body and another 
body may be changed by moving either or both bodies. This 
property is referred to as mobility. Thus, no matter how firm 
the foundation or how heavy and rigid the structure on it, an 
earthquake will move it. The earth itself, the sun, and all the 
stars are moving; there is no such thing as an immovable body. 

12. Inertia. — Inertia means without life, without power of 
self movement. As applied to bodies in a scientific sense, it 
means that if a body is at rest, it cannot put itself in motion; or if 
it is in motion, it cannot bring itself to rest or change its direction 
of motion. This will be explained more fully later. 

13. Weight. — Every body exerts a certain attraction on every 
other body, which tends to make the bodies come together and 
touch one another. As between two bodies of ordinary size, this 
attraction is very small ; but the earth is so large that this attrac- 
tion between the earth and other bodies is very manifest, and the 
result is called weight. Every solid or liquid body, no matter 
how small, will fall to the earth unless sustained by some inter- 
vening body. Gases are also attracted by the earth, even though 
they may rise and go away from it when released. This may be 
proved by weighing an empty vessel in a vacuum (a place where 
there is no air or gas of any kind), then filling the vessel with a gas 
or mixture of gases, say air, and weighing again ; the weight in the 
second case will be greater than in the first case, and the addi- 
tional weight will be the weight of the air. 

14. Indestructibility. — It i s impossible to destroy matter. A 
body may be separated into molecules and the molecules into 
atoms, but the atoms will unite to form other molecules and other 
bodies, and there will be the same number of atoms as before. 
Matter may assume countless forms and undergo innumerable 



6 ELEMENTS OF PHYSICS §4 

changes, but the same number of atoms is present in the uni- 
verse; this is otherwise expressed as the principle of conservation 
of matter, and it means that matter is indestructible. Matter 
may be transformed, but it cannot be destroyed, neither can the 
total amount of matter in the universe be changed. 

15. Specific Properties of Matter. — In addition to the general 
properties, bodies possess certain other properties, not common to 
all bodies, which are therefore called specific properties. Some 
of the most important of these specific properties are rigidity, 
pliability, flexibility, malleability, ductility, tenacity, brittle- 
ness, and hardness. These terms may be defined briefly as 
follows : 

Rigidity is the resistance offered by a body to a change in its 
shape; steel is very rigid. Pliability is the ease with which the 
shape of a body may be changed; lead, copper, etc. are pliable. 
A body is flexible when it can be bent without breaking, as a 
spring, a rope. Malleability is that property that indicates that 
the body possessing it may be rolled or hammered into sheets; 
gold is the most malleable of all known substances. Ductility 
is that property of the body possessing it which indicates that 
the body may be drawn out into wire; platinum possesses greater 
ductility than any other substance. Tenacity refers to the 
resistance offered by some bodies to being pulled apart; steel is 
extremely tenacious. Brittleness is the term used to indicate 
that some bodies are easily broken when subjected to sudden 
shocks; glass, ice, etc. are very brittle. Hardness is that prop- 
erty that indicates that a body possesses the ability to scratch 
some other body. The hardness of any particular body may be 
determined by using it to scratch some other body; it will be 
harder than the body it scratches and softer than any body that 
scratches it. The diamond is the hardest substance known; no 
substance will scratch a diamond except another diamond, in 
which case, either will scratch the other. 



MOTION AND VELOCITY 

16. Definition. — If during a certain space of time, the distance 
between two bodies is increasing or decreasing, one body is said to 
be in motion relative to the other body. If a body A occupy and 
continue to occupy a certain position on the earth's surface, A is 
said to be at rest (or is fixed) relative to the earth; if, now, the 



§4 MATTER, MOTION, AND FORCE 7 

distance between A and another body B continually varies 

during a time t, B is said to be in motion relative to A, or B has 

motion relative to A. 

The use of the word relative in the last paragraph requires a 

special explanation. All motion is relative; there is nothing in 

space that is absolutely at rest. The circumference of the earth 

at the equator is about 25,000 miles, and it turns around on 

its axis once in 24 hours or 86,400 seconds; hence, an object 

on the equator is moving around with the earth at the rate of 

5280 X 25000 ._„., , a rr.. ,. ■ , 

ocfAoo = 1524 leet every second, i he earth is also movmg 

in its orbit about the sun with a speed of about 18 miles every 
second; the sun is also in motion, etc. Consequently, when it is 
stated that a body is at rest, the usual meaning is that the body 
is at rest relative to a point on the earth's surface; and if a 
body is in motion relative to this point, it is moving toward or 
from that point, 

A body may be in motion relative to one point and at rest rela- 
tive to another, both at the same time. For instance, suppose a 
person to be on a slowly moving railway train; if he walk from one 
end of the train to the other at the same rate of speed that the 
train is moving, but in the opposite direction, he will be in motion 
relative to a fixed point on the train, but at rest relative to a fixed 
point beneath him on the earth. In other words, the center of 
his body will remain over the point on the earth until he reaches 
the end of the train. 

17. Path of a Body. — In order to compare motions, it is as- 
sumed that the motion is the same as that of a very small particle 
of the body at the center of gravity of the body; and the line 
described by this particle is called the path of the body. The 
path may be straight or curved, but the length of the path is 
always equivalent to a right line having the same length. If the 
path is a right line, the direction of motion is along that line, i.e., 
along the path ; but if the path is a curve, the direction of motion 
is along a tangent to the path at the point occupied by the body 
(center of the body) at the instant considered. This will be 
explained more fully in the section on mechanics. 

18. Velocity. — Velocity is rate of motion. If the speed of a 
body is uniform, that is, if the body passes over equal distances in 
equal times, the velocity is equal to the distance divided by the 



8 ELEMENTS OF PHYSICS §4 

time. Thus, ii v = the velocity, s = the distance, and t = the 
time, 

s 

The unit of velocity is a compound unit, because it requires 
more than one unit to express it — a unit of distance (length) and a 
unit of time. If s is measured in feet and t in minutes, the unit of 
velocity is one foot divided by one minute, and is called one foot 
per minute, the word per meaning divided by; whenever the word 
per occurs in the name of a compound unit, it always has this 
meaning. If s is measured in miles and t in hours, the unit of v is 
one mile per hour; and if s is in centimeters and t in seconds, v is in 
centimeters per second. 

19. When a body passes through equal distances in equal 
times, its velocity is said to be uniform or constant otherwise it is 
variable. But whether uniform or variable, the velocity of a 
body at any instant is the distance (length of path) it would 
travel in a unit of time if the velocity it had at the instant con- 
sidered were uniform. For instance, suppose the speed of a body 
were such that at the instant it arrived at a certain fixed point, it 
would travel 852 feet in one minute if it continued at the same 
rate of speed for one minute; then the velocity at the point 

Q CO 

considered is 852 feet per minute, or -^ = 14.2 feet per second. 

If the velocity in miles per hour were desired, note that if the body 
kept up the same rate of speed for one hour = 60 minutes, it 

would travel 852 X 60 = 51,120 feet = ^^^ = 9.6tt miles; 

hence, the velocity at the instant considered is 9.6it: miles per 
hour. 

20. Acceleration. — Acceleration is rate of change of velocity. 
For example, if the velocity of a body is not uniform, it either 
increases or decreases, and the rate of increase or decrease is 
called acceleration, when the rate of change is uniform, the accel- 
eration is also uniform. Thus, suppose the velocity of a body at a 
certain instant were 120 feet per second, and 4 seconds later the 
velocity were 180 feet per second. If the gain in speed were 
constant, the acceleration would also be constant, and the gain 

in velocity would be at the rate of -z ^ ^^ ^^®* P^^ 

second every second; this would be expressed as 15 feet per 



|4 MATTER, MOTION, AND FORCE 

second per second, or 15 ft. per sec.^, which latter form means 
that the time in seconds is to be squared. The unit of accelera- 
tion is almost invariably taken as one foot per second per second 
= 1 ft. per sec.^ or as one centimeter per second per second = 1 
cm. per sec.^ 

If, in the case just mentioned, the velocity had decreased 
uniformly from 120 ft. per sec. to 80 ft. per sec, the acceleration 

would have been j = — 10 ft. per sec. 2, the acceleration in 

this case being negative. 

To understand clearly the reason for squaring the time in the 
unit of acceleration, perform the calculation by using the 

80«._120i*^ 
units in connection with the numbers. Thus, - 



4 sec. 



4 sec. sec. 2 



— 10 — '— = — 10 ft. per. sec. 



FORCE, MASS, AND WEIGHT 

21. Force. — By reason of its inertia, a body cannot put itself 
in motion or bring itself to rest or change its shape ; to do any of 
these things, the body must be acted upon by some force. There- 
fore, force may be defined as that which tends to change the state 
of rest or motion of a body or which acts to produce deformation 
(change of shape) of a body. In the case of a gas, the force due 
to the motion of the molecules causes the body to expand and 
fill the vessel that contains it; in the case of a liquid, the force 
due to the earth's attraction causes the liquid to spread and con- 
form to the shape of the vessel holding it, and makes the upper 
surface flat. 

Forces are called by various names, according to the effects 
they produce, some of them being : attraction, repulsion, adhesion; 
cohesion, gravitation (or weight), action, reaction, friction, etc. 
But, whatever the effect produced, every force is equivalent to 
a pull or a push. A pull usually tends to elongate (stretch) the 
body on which it acts, and a push usually tends to compress 
(shorten) the body on which it acts. 

22. Cohesion and Adhesion. — Cohesion is the force that holds 
molecules together to form a body; it is an attractive force 



10 



ELEMENTS OF PHYSICS 



§4 



between like molecules. The force of cohesion is extremely- 
weak in the case of gases and extremely strong in the case of 
solids. When two bodies are held together by the attraction of 
unlike molecules, the attracting force is called adhesion; thus, 
two pieces of wood may be held together by glue, and the mole- 
cules of glue being different from those of wood, the force holding 
the bodies together is called adhesion. Likewise grease adheres to 
iron, wax to paper, etc. 

23. Measure of Force. — The force with which the earth 
attracts bodies on or near its surface is called the attraction 
(or force) of gravity or, more simply, gravity or weight. Weight, 
then, is the force (pressure or push) exerted by a body resting 
on the earth or on another body. If the body be suspended from 
another body, the connection being a flexible body, say a string 
or a spring, it will exert a pull on the second body and will 
stretch the string or spring. In either case, the effect of the 
force will be exactly the same. The force of gravity 
is, therefore, very convenient for comparing forces. 

The force of gravity always acts in the direction 
of a line drawn from the center of gravity of the body 
to the center of the earth ; this line is called a vertical 
line, and any line perpendicular to it is a horizontal 
line. It will thus be seen that any line (imaginary) 
drawn on the upper (free) surface of a liquid is a 
horizontal line. A vertical line is also called a plumb 
line, because if a plumb bob be suspended from a 
string, the centerline of the string will point to the 
earth's center when the plumb comes to rest; hence, 
this line is vertical. 

24. Fig. 1 shows a spring balance, from the hook 
of which is suspended a standard one-pound weight; 
the indicator then points to the 1 mark on the 
scale. If, now, the one-pound weight be replaced 
by another body that brings the indicator to the 1 
mark, the force exerted in both cases is the same. Con- 
sequently, the force with which either body presses against 
any other body on which it rests is called one pound. It is 
evident that a weight of one pound will exert a force (pressure) of 
one pound when resting on another body; consequently, forces 
are measured by their equivalent in weights, and are expressed in 



Fig. 1. 



§4 MATTER, MOTION, AND FORCE 11 

pounds, grams, or kilograms. For instance, the force exerted by 
a hammer in driving a nail a certain distance into a board would 
be measured by the weight of a body that, resting on the nail, 
would press it the same distance into the wood in the same time; 
the force with which steam presses against a piston of an engine 
is equal to the weight that would exert an equal pressure on 
the piston. Observe that both weight and force are measured 
with a spring balance, the measure of either being determined 
by the amount that the spring is stretched, as indicated by 
the scale. 

25. Mass. — By mass is meant the amount of matter contained 
in a body. A little consideration will show that the mass of a 
body cannot be determined by measuring its volume. A cubic 
foot of spruce wood will weigh, say, 35 pounds, while a cubic foot 
of cast iron will weigh 450 pounds; it is evident, therefore, that a 
cubic foot of cast iron contains more matter than a cubic foot of 
spruce. In other words, the mass of cast iron is almost 13 times 
as great as the mass of an equal volume of spruce wood. Conse- 
quently, the mass of a body is determined by weighing it; but 
the mass is not equal to the weight, as will now be shown. 

26. If a body be allowed to fall freely in a vacuum at a place 
whose latitude is about 41°, or very nearly that of New York, the 
uniform acceleration will be 32.16ft. persec.^ The body will have 
this acceleration no matter what its weight, whether it be 1 ounce 
or 1 ton, because it makes no difference whether the body be di- 
vided into 2000 X 16 = 32000 equal parts and all fall separately 
or whether they be joined together and all fall as one body. The 
force which causes the body to fall and gives it the acceleration it 
receives is the force of gravity, which is equal to the weight of the 
body, and this is a constant (uniform) force acting on the body 
during its time of falling. It is therefore plain that if a body 
weighing one pound receive an acceleration of 32.16 ft. per sec.^ 
when falling freely, a force of one pound acting on a body weighing 
one pound and free to move will give it the same acceleration. It 
would also seem as though a force of one pound acting on a body 
weighing 2, 3, n times as much, that is, 2, 3, n pounds, would 

receive an acceleration of only |, ^d, - th as much, and experi- 
ment shows this to be the case. Therefore, if / = the force 
acting on a body, w = the weight of the body, a = the accelera- 



12 ELEMENTS OF PHYSICS §4 

tion produced by the action of the force /, and g = the accelera- 
tion produced by gravity, 

f : w = a : g 

. wa w ^^ 
J = — = — X a = ma, 

g g 

w 
when m = — • To find the value of w, it must be found by weigh- 
ing the body with a beam scale instead of a spring balance (in 
order that standard weights may be used; the spring balance gives 
true weight only under specific conditions) ; the value of g may 
be found by direct experiment or by calculation; then, knowing 
the value of g for any locality and weighing the body under 
consideration on a beam scale, the value of m can be found 
by substituting the known values in the formula. The value 
thus found for m is called the mass of the body. From the pre- 
ceding equation, / = ma, that is, the force which will give to any 
body an acceleration a is equal to the mass of the body multiplied by 
the acceleration. 

27. The reason for defining the mass of a body as its weight at 
any particular place divided by the value of g at that place is 
that the value of g is different for different places on the earth's 
surface; it is least at the equator and greatest at the poles, be- 
cause the earth is not a perfect sphere, a point on the surface 
at the poles being nearer the earth's center than a point on the 
surface at the equator. The farther the body is from the center 
the smaller is the value of g; consequently, the value of g is less 
at the top of a high mountain than at sea level. The weight 
of a body varies directly as the value of g, but the mass, which 
measures the quantity of matter remains constant. Letting w' 
and w" (read w prime and w second) be the weights of a body at 
two different places, m the mass of the body, and g' and g" the 
values of the acceleration produced by gravity at those places, 

w' : w" = g' : g" 

w' w" 
or w'g" = w"g', from which -7 = — ^ = m. This last equation 

shows that w is constant, since if g decreases, w also decreases, 
and if g increases w also increases, both in the same proportion. 

28. The C. G. S. System.— In what is called the C. G. S. 
system (C. G. S. is the abbreviation for centimeter-gram-second), 
the value of g is expressed in centimeters per sec^. For latitude 



§4 MATTER, MOTION, AND FORCE 13 

45°, which is half way between the equator and the poles, 
g = 980.665 cm. per sec. 2, very nearly, and this value of g is 
now quite generally accepted as the standard by scientists. Then, 
since 1 cm. = .03280843 ft., 980.665 cm. = 980.665 X .03280843 
= 32.1741 ft. For latitude of New York, g = 980.223 cm. per 
sec.2 = 32.1599, say 32.16 ft. per sec^. 

29. Weight. — As previously stated, weight is caused by the 
attraction of the earth. This attraction is mutual, the body 
attracting the earth as much as the earth attracts the body. The 
mass of the earth is so great, however, that the distance the earth 
moves is too small to be measured. The equatorial diameter of 
the earth is about 26.4 miles greater than the polar diameter; 
in other words, a person standing at the pole (north or south pole) 
is over 13 miles nearer the center than when he stands on the 
equator. The nearer a body is to the center the greater is the 
attractive force (force of gravity), the greater is the value of g, 
and the greater is the weight of the body when measured with 
a spring balance. If, however, a body is weighed with a beam 
scale, the weight will be the same anywhere, because the weight of 
the counterpoise changes in the same proportion as the weight of 
the body. For example, if a barrel of sugar weighs 300 pounds at 
New Orleans on a beam scale, it will weigh the same at New York, 
at London, or at the pole, using the same scale. But if a spring 
balance be used that has been graduated in accordance with the 
value oi g = 980.665 cm. per sec.^ = 32.1741 ft. per sec.^, the 
weight recorded at New Orleans will not be the true weight. 
The value of g for New Orleans is 32.1303 ft. per sec.^ The 
true weight is the weight that would be recorded at a place where 
the value of g is 32.1741 ft. per sec.^ and may be found from the 
proportion of Art. 27, or 300 : w; = 32.1303 : 32.1741, from which 
w — 300.409 pounds. This is the weight that would be recorded 
on a beam scale, if the weight of the counterpoises were stand- 
ardized to correspond with g = 980.665 cm. per sec.^ = 32.1741 
ft. per sec. 2 

In commercial transactions and in ordinary engineering calcu- 
lations, no attention is paid to the variation in the value of force 
(or weight) due to difference in latitude; but in all scientific 
investigations requiring accuracy in results, these variations 
must be considered. For all practical purposes, the value of any 
force at any place may be considered as the equivalent of the 
weight that Avill produce the same effect at that place. 



14 ELEMENTS OF PHYSICS §4 

30. Density. — Density is the quantity of matter contained in a 
unit of volume; it is the mass of a unit of volume. If the density 
of a body be represented by D, the mass by m, the volume by V, 
the weight by w, and the acceleration due to gravity by g, 

^ V gV ^^^ 

Solving this equation for m, 

m = DV (2) 

that is, the mass of any body is equal to the product of its density 
and volume. 

Density is also frequently defined as the weight of a unit of 
volume, in which case, 

C=^ (3) 

The density of gases is almost always expressed as in the second 
of the above definitions. If the unit of weight be taken as 1 
pound and the unit of volume as 1 cubic foot, the density of a 
body by the second definition is in pounds per cubic foot, and the 
density may always be found by weighing 1 cubic foot. Another 
name for the weight of a unit of volume is specific weight; thus, 
if the weight of a cubic foot of cast iron is 450 pounds, its specific 
weight is 450 pounds per cubic foot, which is also the density in 
accordance with the second definition. In accordance with 

450 
the first definition, the density of cast iron is oo 1741 y 1 

13.98 — . There is no name for the unit of density as determined 
by formula (1), which may be called the specific mass; neither is 
there any for the unit of mass. Consequently, for the want of a 
better name, these units may be called density units and mass 
units, respectively. 



WORK AND ENERGY 

31. Work. — That point of a body at which a force is applied or 
at which it may be considered as acting is called the point of 
application. If, as the result of the action of a force, the point of 
application is moved through a certain distance, work is done. 
Thus, if a body weighing 15 pounds is lifted 6 feet vertically, work 
is done in doing this. The unit of work is usually taken as a. foot- 
pound; this is a compound unit, and means that a resistance of one 
pound has been overcome through a distance of one foot, and is 



§4 MATTER, MOTION, AND FORCE 15 

exactly equivalent to 1 pound X 1 ft. In other words, to find the 
work done, multiply the force {in pounds) by the distance (in feet) 
through which it acts. The work done in raising 15 pounds through 
a vertical distance of 6 ft. is 15 X 6 = 90 ft. -lb. 

Suppose the average pressure on the piston of a steam engine is 
63 pounds per square inch, that the diameter of the piston is 
15 inches, and that it moves 18 inches during one stroke; then the 
work done per stroke may be found as follows. The total average 
pressure during the stroke is 63 times the area of the piston in 
square inches, since the average pressure is 63 pounds per sq. in. ; 

IT 

hence, total pressure = 63 X t X 15^. The distance through 

18 
which this pressure (force) acts is 18 in. = t?; = 1.5 ft. Conse- 
quently, the work done is 1.5 X 63 X .7854 X 15^ = 16,700 
ft.-lb. 

32. Work is the overcoming of a resistance through a distance; if a 
force acts on a body without moving the point of application, no 
work is done. For instance, if a man try to lift a stone that is too 
heavy for him to move, he will apply considerable force to the 
stone, but no work will be done on the stone. In the case of a 
falling body, gravity is the acting force, and equals the weight of 
the body; the work done is the weight of the body multiplied by 
the vertical distance through which the body falls. 

Observe that the work done is not dependent on the time it 
takes to do it. Thus, 15 pounds raised through a vertical dis- 
tance of 6 feet is equal to 15 X 6 = 90 ft.-lb. of work; and it 
makes no difference whether it took one second to raise the 
weight or one month, the work done is the product of the acting 
force and the distance through which it acts. It should also be 
noted that when gravity is the acting force, it makes no difference 
how the body gets from one level to the other (that is, what the 
shape of the curve representing its path), the work done is the 
weight of the body multiplied by the vertical distance (perpendic- 
ular distance) between the level that includes the starting point 
and .the level that includes the stopping point. Thus, in hauling 
a wagon up a hill, suppose the weight of the wagon and its load 
is 900 pounds and that the difference of level between the top of 
the hiU and the bottom is 120 feet; then, neglecting friction and 
other resistances, the work done is 900 X 120 = 108,000 ft.-lb., 
regardless of how the top of the hill was reached. The result is 
the same as though the wagon had been lifted bodily 120 ft., 



16 ELEMENTS OF PHYSICS §4 

as by an elevator. This is because when the wagon moves up the 
hill, the acting force is not equal to the weight of the body, but 
is considerably less, depending on the slope. When the body 
moves vertically up or down, then the acting force is equal to the 
weight. 

33. Energy. — Energy means capacity for doing work — ability 
to do work. A body in motion cannot instantly be brought to 
rest; to bring it to rest, a force must act through a distance, no 
matter how short. For instance, when a blow is struck with a 
hammer, a dent is made in the body struck, and the depth of this 
dent is the distance passed through by the hammer in coming to 
rest. This distance (in feet) multiplied by the force of the blow 
is the work done. At the instant the hammer strikes, but before 
any work is done, the hammer has capacity for doing the work 
that is done as the result of the blow, and this capacity for doing 
work is called energy. 

Energy is measured in the same units as work, that is, in foot- 
pounds. Suppose a body weighing 500 pounds to fall through a 
vertical height of 16 ft.; the work it can do is exactly equal to the 
work required to be done on the body to raise it through a vertical 
height of 16 ft., neglecting resistance of the air, friction, etc. in 
both cases. This work is 500 X 16 = 8000ft.-lb.; consequently, 
the energy of the body is 8000 ft. -lb. also. 

34. Kinds of Energy. — It is customary to divide energy into 
two classes — potential energy and kinetic energy. Potential 
energy is that due to the position of a body. Consider, for 
example, a pile driver. "When the hammer that drives the pile 
has been raised to its highest point and is ready to be released, it 
possesses energy due to its position, the amount being equal to the 
weight of the hammer in pounds multiplied by the height in feet 
of the hammer above the pile, which is equal to the vertical 
distance between the bottom of the hammer and the top of the 
pile. This energy is potential energy, so called because it may or 
may not be used. Potential means possible; hence, potential 
energy means possible energy. 

Kinetic energy is actual energy — it is the energy of a body in 
motion. In the previous illustration, when the hammer is 
released, it falls and the potential energy is gradually changed 
into kinetic energy until, at the instant the hammer hits the top 
of the pile, all the potential energy has been changed into kinetic 



§4 MATTER, MOTION, AND FORCE 17 

energy. The sum of the potential energy (if any) and the 
kinetic energy (if any) is the total energy. Thus, when the 
hammer has fallen through Kth the height, 3'^th of the potential 
energy has been changed to kinetic energy; hence, the total 
energy is made up of ^^th of the original potential energy and the 
other 3-^th is kinetic energy. Again, when a rifle is fired, the 
potential energy of the powder is changed to kinetic energy in the 
barrel of the gun; the kinetic energy is converted into work on the 
bullet, and when the bullet leaves the gun, it has kinetic energy, 
which is converted into work when the bullet is stopped. It will 
be noted that the bullet does not have potential energy at any 
time; it has either no energy at all or it has kinetic energy. If, 
however, the bullet be fired vertically upward, and it meet with no 
resistance from the air, the action of gravity causes the velocity 
to decrease, until it stops, but immediately begins to fall. Dur- 
ing the rise, the kinetic energy has been changing into potential 
energy, and at the instant it begins to fall, all the kinetic energy 
has been changed to potential energy. As it falls, the potential 
energy decreases and the kinetic energy increases; and when 
it again reaches the earth, all the potential energy has been 
changed to kinetic energy, equal in amount to the original kinetic 
energy it had when it left the gun. 

All work is the result of kinetic energy, and potential energy 
of some kind must be converted into kinetic energy before any 
work can be done. In the case of the gun, the potential energy 
was in the powder; in the case of a steam engine, the potential 
energy is in the steam; and in the case of the pile driver, the 
potential energy was in the position of the hammer before it was 
released. Potential energy is always stored energy, while kinetic 
energy is active energy that is, or can be, transformed into work. 

Energy exists in innumerable forms; like matter, it cannot be 
destroyed, but may be changed in various ways. The energy 
stored in coal was received from the sun millions of years ago; 
it is changed into heat (a form of kinetic energy), by combus- 
tion, and generates steam; steam drives the engine, which, in 
turn, drives a dynamo that generates electricity; thus energy that 
originally came from the sun is converted into electric energy, 
used to drive motors, light lamps, etc. The energy stored in 
food is converted into muscular energy, and enables us to five, 
move, work, and play. 



18 ELEMENTS OF PHYSICS §4 

HYDROSTATICS 



PASCAL'S LAW 



FLUID PRESSURE 

35. Equilibrium. — A body is said to be in equilibrium when 
it is at rest under the action of forces or if in motion, there is no 
change in its velocity or direction of motion. A block of stone 
resting on the earth is in equihbrium; the forces acting on the 
stone are the force of gravity, which acts downwards and pulls 
the block toward the center of the earth, and the pressure of the 
earth against the block, called the reaction of the earth, which 
acts upwards and prevents the block from moving toward the 
center of the earth. A stick balanced on a knife edge is in 
equilibrium ; gravity tends to pull the part of the stick on one side 
of the knife edge one way and the other part the other way, the 
two balancing each other; but the total pull toward the earth is 
the weight of the stick, and this is resisted by the reaction of the 
knife edge, which acts upward. A person riding- on a railway 
train that is moving in a straight line, along level ground, at a 
constant rate of speed is in equilibrium; the force exerted by the 
locomotive in pulling the train just balances that required to 
overcome the resistance of the air, friction, etc. ; but, if the train 
commences to go around a curve or up or down a grade, the 
person will no longer be in equilibrium — ^the direction of motion 
has been changed; or, if the speed of the train is increased or 
decreased, he will no longer be in equilibrium — the velocity has 
been changed. If the change is sudden, he will be thrown for- 
wards, backwards, sideways, up, or down, according to the nature 
of the change in motion. 

Hydrostatics is that branch of science which treats of fluids at 
rest under the action of forces, that is, of fluids in equilibrium; 
the word is derived from the Greek, and literally means water 
at rest, hydro meaning water and statics meaning standing or at 
rest. Hydrostatics is usually limited to water and other liquids, 
but may also be applied to gases. 

36. Pressure. — When a steady force having the effect of a 
push acts on a body, it is called a pressure. When the force 
acts very suddenly and for a short time, it is called an impulse, 



§4 HYDROSTATICS 19 

or a blow; but, if desired, any blow may be considered as a pres- 
sure acting for a very short time. 

Suppose a log having a mean diameter of 16 in., 20 ft. long, 
with ends at right angles to the axis, and weighing 960 pounds to 
stand on one end; it will exert a pressure of 960 pounds on the 
surface that supports it, and this pressure will be evenly distrib- 
uted over a surface whose area is equal to the area of the end of 
the log. Since the log is not exactly round, the area of the end 
may be considered as equal to the area of a circle whose diameter 
is the same as that of the log, but taking t equal to 3 instead of 

3 
3 . 1416. Hence, the area is j X 16^ = 192 sq. in. The pres- 
sure exerted by the log on each square inch is then Jq^ ^ ^ pounds 

per square inch. When a pressure is stated as a force per unit of 
area, it is called specific pressure, and the pressure on the entire 
area is called the total pressure. In the above case, the total 
pressure was 960 pounds, and the specific pressure was 5 pounds 
per sq. in. 

37. Specific pressures are generally stated as pounds per square 
inch or pounds per square foot, but in connection with fluid 
pressure, they may be stated as feet of water, inches of mercury, 
or atmospheres, terms that will be explained later. In the metric 
system, specific pressures are usually expressed in kilograms per 
square centimeter or kilograms per square meter; meters of water 
and millimeters of mercury are also used. 

38. Total pressures may be considered as distributed over an 
area or as acting on a line passing through the center of gravity 
of the area pressed against. Thus, in the case of the log standing 
on one end, the total pressure of 960 pounds may be considered as 
distributed over an area equal to the area of one end or as a 
concentrated force that acts at the center of gravity of the area. 
Insofar as any movement of the body under the action of the 
force (pressure) is concerned, the effect will be the same in either 
case. 

39. Transmission of Pressure. — When a force (pressure) acts on 
a solid, it is transmitted through the solid in a line that forms an 
extension of the line of action; thus, if one end of a chisel be hit 
with a hammer or mallet, the other end of the chisel will impart 
to any body it touches a blow of the same force as that which was 
received by the chisel. In other words, the chisel transmitted the 



20 



ELEMENTS OF PHYSICS 



§4 



480 lbs. 



W 



blow it received at one end to the other end. The blow, itself, 
measured as a force, may be different at the two ends, by reason 
of a difference in the shapes of the ends, and the material of the 
body struck by the chisel will probably be different from that of 
the chisel; but all the energy received at one end of the chisel will 
be transmitted undiminished to the other end, provided the blow 
is not hard enough to injure (deform) the chisel in any way. 
There will be no pressure in any direction except lengthwise of 
the chisel. The same is true of a weight resting on a pillar; the 
pressure due to the weight is transmitted from one end of the 
pillar to the other, and the total pressure is the same at either 
end. 

In the case of fluids, the result is entirely different, because the 
molecules of the fluid are free to move in any direction. When 
pressure is applied to a solid that is not too pliable, it retains its 

shape for all practical purposes, and 
any movement of the molecules must 
be lengthwise only. There is a cross- 
wise (lateral) movement, but it is so 
slight that for ordinary pressures, it 
may be neglected, and the lengthwise 
(longitudinal) movement may also be 
neglected. 

40. When a fluid is subjected to 
pressure, the pressure is transmitted 
undiminished in every direction. It 
is to be noted that the word pressure 
in this case means specific pressure. 
Referring to Fig. 2, which represents 
a vertical longitudinal section passing 
through the axis of a cylindrical 
vessel, suppose the vessel to be filled 
with a fluid, say water, and fitted 
with a tight-fitting piston P, which 
rests on top of the water. Suppose further that a weight W 
of 480 pounds is placed on top of the piston, as shown. The 
pressure on top of the water due to the weight W will be 480 
pounds and this will also be the pressure on the bottom of the 
vessel (neglecting the weight of the water), the result being 
the same as though the space occupied by the water were a 
round block of wood. If the area of the bottom pf the vessel is 



^^/y/-///////.- 



'///////'y/y///y/'''.'/////.''X/// 



Area 60 sq, in. 

Fig. 2. 



§4 



HYDROSTATICS 



21 



60 sq. in., the specific pressure is -^ = 8 pounds per sq. in., and 

this specific pressure is transmitted in every direction, as indicated 
by the arrows — upward, downward, laterally, and at any angle 
whatever. 

Had the vessel been filled with a gas instead of water, the same 
result would have been obtained, the only difference being that 
water is only very slightly compressible and the change in volume 
is not noticeable; but gas is so compressible that the volume 
would be very much smaller. For this reason, water will be 
adopted as the fluid under consideration in what follows, unless 
otherwise specified ; but what is true of water in general is also 
true of any other liquid and of gases. 

41. Pascal's Law. — The law governing the transmission of 
pressure in fluids confined in a closed vessel was discovered by 
Pascal, a famous French scientist 
and mathematician (1623-1662), 
and is known as Pascal's law; it 
may be stated as follows: 

Specific pressure exerted on any 
fluid confined in a closed vessel is 
transmitted in all directions and 
acts on all surfaces touched hy the 
fluid in a direction perpendicular 
to those surfaces. 

42. If the surface pressed 
against by the fluid is a curved 
surface, the perpendicular to 
the surface at any point is 
called a normal, and a normal 
to a curve at any point is per- 
pendicular to the tangent to 
the curve at that point. This 
is illustrated in Fig. 3, where 
ECFG represents a curved surface, and the arrows a, b, c, d, 
etc. represent normals to this surface. Let n represent the 
center of gravity of the surface, and let P represent a pressure 
which, acting on the surface at n in the direction of the normal 
N will produce the same effect on the surface as is produced 
by the pressure of the fluid. To flnd the value of P, let RS be a 
plane perpendicular to the normal N, and let A'B'C'D' be the 




Fig. 3. 



22 ELEMENTS OF PHYSICS §4 

projection of ECFG upon the plane RS. Let a = the area of 
A'B'C'D' and p = the specific pressure exerted on the fluid; then 

P = va, 

that is, the total normal pressure against any surface is equal to 
the specific pressure multiplied by the projected area of the sur- 
face, projected on a plane at right angles to the normal drawn 
through the center of gravity of the surface. This total normal 
pressure tends to move the entire surface under consideration in 
the direction of the normal. In the case of the cylinder, Fig. 

2, the diameter of the cylinder is d = ^ /— = -» /- — — — = 8.74 

\7r \3.1416 

in., nearly. Assuming that the depth of the water in the cylin- 
der is 25 in., the projected area of one-half the cylinder on a plane 
perpendicular to the normal through the center of gravity of the 
half-cylinder (which will be on a line midway between the top 
and bottom and midway between the parallel sides of the half- 
cylinder) will have the shape of a rectangle whose length is 
25 in. and breadth is 8.74 in. The projected area will then be 
25 X 8.74 = 218.5 sq. in. Since the specific pressure is 8 lb. 
per sq. in., the total normal pressure on the half-cylinder is 
218.5 X 8 = 1748 pounds. This same pressure acts on the other 
half of the cylinder, but in the opposite direction. Therefore, 
the total lateral pressure due to the weight of 480 pounds tending 
to separate one-half of the cylinder from the other half is 1748 
pounds or almost four times the weight, in this case. 

43. Figure 4 represents a flask filled to the line mn with water, 
the space between mn and the bottom of the plunger p being 
filled with air. As the handle H is pushed down, the air is com- 
pressed, being confined between the top of the water and the 
bottom of the plunger, and the greater the pressure the smaller 
becomes the volume of the air. The pressure is transmitted in 
all directions through the air, from the air to the water, and in 
all directions through the water. Assuming that the pressure 
on H is such that the pressure on the confined air is 12 pounds 
per square inch, this specific pressure will be transmitted to all sur- 
faces touched by the water. Little pistons are placed in the 
vessel at a, h, c, etc., and their cross-sections have the following 
areas: a = 1.5 sq. in., h = 2.75 sq. in., c = 2.25 sq. in., d = 3.5 
sq, in., e = 4.75 sq. in., / = 2.5 sq. in., g = 5.5 sq. in., and 
A = 3 sq. in. Then, neglecting the weight of the water, the 



§4 



HYDROSTATICS 



23 



forces acting on the inside faces of the pistons, tending to force 
them out, and which must be resisted by an equal and opposite 
force on the other side of 
the pistons, indicated by 
the arrows, are respectively 
1.5 X 12 = 18 pounds, 2.75 
X 12 = 33 pounds, 2.25 X 
12 = 27 pounds, 3.5 X 12 
= 42 pounds, 4.75 X 12 - 
57 pounds, 2.5 X 12 = 30 
pounds, 5.5 X 12 = 66 
pounds, and 3 X 12 = 36 
pounds. These are all 
normal pressures; the pis- 
tons are supposed to be 
round, but the faces may 
be flat or curved, the areas 
given above being the pro- 
jected areas, and equal to 
the squares of the diame- 
ters of the pistons multi- 
plied by .7854. 

44. Hydrostatic Ma- 
chines.^ — ^A hydrostatic 
machine is a device for 
raising heavy loads or exerting great pressure as in pressing 
bales of pulp. The principle governing their operation is 
illustrated in Fig. 5. P and p are pistons working in the cyl- 




FiG. 4. 



EZM-P 



Fig. 5. 



inders A and B. P carries a weight represented by W, and the 
force pushing p down is represented by the weight w. The space 



24 ELEMENTS OF PHYSICS §4 

beneath the pistons is filled with water, oil, or other liquid, and 
the two cylinders are connected by a pipe a, so that the liquid 
can flow freely from one cylinder to the other. As p moves down 
the liquid in A flows through a into B, causing P to rise and with 
it the weight W. The specific pressure exerted by y is trans- 
mitted to P. Let d = diameter of p and D = diameter of P; 
then .7854 d^ = area of p and .7854 D^ = area of P, and the speci- 

w 
fie pressure exerted by w is 70^4^2 - -^^^ *^i^ must be the same 

W 

as the specific pressure exerted on W, which must be rjonAn^ 

w TV w W 

therefore, ^354^2 = 7354^)2 ^ ^^ ^ ^ d^> ^^^^ which 

That is, the weight that can be lifted (or the pressure that 
can be exerted) by the large piston is equal to the product of 
the weight (or force exerted) on the small piston and the square 
of the diameter of the large piston divided by the square of the 
diameter of the small piston. 

Example. — If the diameter of the small piston in Fig. 5 is 1}4 in-; of the 
large piston 12 in., and the force exerted on the small piston is 38 lb., what 
weight W can be raised by the large piston? 

Solution. — Substituting in the above formula the values given, 
Qs V 192 
W = \l2 = 2432 pounds. Ans. 

45. Attention is called to the fact that the work done on the 
small piston is (neglecting friction and other hurtful resistances) 
exactly the same as the work done on the large piston. Let h = 
the distance (height) passed through by the small piston and 
H = the distance passed through by the large piston ; then wh = 

wh 
WH, from which H — ^- If, in the example just given, w 

38 X 2 
moves 2 in., W will move only „.„t^ = .03125 in. In other 

words, what is gained in pressure or lifting force is lost in speed. 
This is a general law, and is true of any machine. The converse 
is equally true; that is, if there is a gain in speed, there is a loss 
in the force that can be applied. For example, suppose that, 
referring to Fig. 5, it were desired to move the small piston upward 
6 in. while the large piston was moving downward 3^^ in. ; if the 
force W exerted on the large piston is 4000 pounds, what pres- 



§4 HYDROSTATICS 25 

sure will be exerted on the small piston? Since wh = WH, 

wX 6 = 4000 X .5 = 2000, and w = — g- = 333^ pounds. 

In other words, a force of 4000 pounds moving J-^ inch can raise 
a weight of only 333| pounds through a height of 6 inches, and 
what is gained in speed (distance) is lost in applied force. 

46. The conclusion arrived at in the last article may also be 
verified as follows: Taking the dimensions of the example in 
Art. 44, when the small piston moves down, it forces an amount 
of liquid into the cylinder B that is equal to the area of the small 
piston multipUed by the distance through which it moves; that 
is, it is equal to the volume of a cylinder whose diameter is the 
same as that of the piston and whose altitude is equal to the 
distance that the piston moves. The volume of the liquid under 
the large piston is increased the same amount, and is equivalent 
to a cylinder whose diameter is that of the large piston and 
whose altitude is the distance the piston is raised, or H. Con- 
sequently, assuming that piston p moves 2 inches, .7854 X 1.5^ 

X 2 = .7854 X 122 x H,orH = ^^^^ = j|o ^^' ^ '^^^^^ 
in., and what is gained in applied force is lost in speed. Observe 
that this result is exactly the same as that obtained in Art. 45. 

47. Pressure of Liquid Due to its Own Weight. — In what has 
preceded, only the effects produced by the application of 
an external pressure have been considered. Since liquids have 
weight, they exert pressure on the vessels containing them, and 
the methods of determining this pressure will now be considered. 
Referring to Fig. 2, suppose the piston and its weight of 480 
pounds to be removed; the total pressure on the bottom of the 
cylinder will then evidently equal the weight of the liquid, and 
the specific pressure on the bottom will equal the weight of the 
liquid divided by the area of the bottom, and will be transmitted 
downward, upward, and laterally. This specific pressure will 
not, however, be the same for all parts of the liquid, since the 
pressure on the top of the liquid will be 0. At any point be- 
tween the top of the liquid and the bottom of the vessel, the 
pressure will be equal to the weight of the liquid corresponding 
to the depth of the point below the top of the liquid. Insofar as 
the downward pressure is concerned, the case is exactly analo- 
gous to a pile of bricks, see Fig. 6. The pressure on the surface 
AB that supports the bricks is equal to the weight of all the 



26 



ELEMENTS OF PHYSICS 



§4 




Fig. 6. 



bricks, or 7 bricks in this instance. There are 6 bricks on top 
of brick No. 1, and the pressure on the top of this brick is 
equal to the weight of 6 bricks, the depth of the pile between 
brick 1 and the top of brick 7. The pressure on top of brick 4 
is 3 bricks, the number of bricks between the top of brick 4 
and the top of brick 7; etc. 

Suppose the size of the bricks to be 4 in. wide, 8 in. long, and 
2 in. thick, and that each brick weighs 4.5 pounds. The total 

pressure on the surface AB is 7X 4.5 
= 31.5 pounds, and the specific 
pressure is 31.5 -^ 4x8 = .984375 
pound per square inch. The weight 
of 1 cubic inch of brick is 4.5 -^ (4 X 
^ 8X2)= .0703125 pound. The depth 
i" of the pile is 7 X 2 = 14 in., and 
.0703125 X 14 = .984375 pound = the 
weight of a prism of brick 1 in. square 
and 14 in. high. But this value is 
the same as the specific pressure; 
hence, the specific pressure for any 
depth is equal to the weight of a prism of the brick whose base 
is the unit of area and whose height is the depth of the brick 
at the point considered. This also applies to fluids. 

48. A cube of water measuring 1 ft. on each edge (1 cubic foot) 
weighs 62.4 lb. at its temperature of maximum density (4°C. or 
39.2°F.) ; it weighs less at higher temperatures, but. unless very 
exact results are desired, the weight of water may be taken as 
62.4 pounds per cu. ft. A column of water 1 in. square and 1 ft. 
high evidently weighs 62.4 -^ 144 = i| = A}i pound. There- 
fore, for water at any depth, let p = the specific pressure and 
h = the depth in feet; then, 

p = Alh = Uh (1) 

If the depth be taken in inches, the weight of 1 cu. in. = 62.4 

-^ 1728 = ^Vo = .036111 pound, say .0361 pound for practical 

purposes. Letting h' be the depth in inches, the specific pressure is 

P = iM' = -0361^' (2) 

If it be desired to find the depth necessary to produce a given 

specific pressure, solve the above formulas for h and h', obtaining 

h ^ ^p = 2.3077p (3) 

and 



h' = \^p = 27.6922? 



(4) 



§4 



HYDROSTATICS 



27 



For practical purposes, a depth of 2.31 ft. or 27.7 in. of water 
may be considered as equivalent to a pressure of 1 lb. per sq. in. 

49. Pressure Due to Liquid on Submerged Surface. — In Fig. 7, 
let ahcd be a flat plate submerged in the water contained in the 
tank T. It is evident that the pressure (specific) at the edge ah is 
less than it is at the edge dc. Let / be the center of gravity of the 
plate; then the specific pres- 
sure at / will be the same at 
any point along a horizontal 
line drawn on the plate and 
passing through /, because 
every point on such a line 
will be at the same depth 



below the level of the liquid. ;:!H:i'ii^i^ 




Fig. 7. 



Further, this specific pressure 

will be the average pressure 

on the plate, and when multiplied by the area of the plate, the 

product will be the total normal pressure P on the plate. Let 

h = the depth of / below the surface of the liquid, a = area of 

plate, and P = the total normal pressure on the plate; then, 

P = ahw (1) 

in which w is the specific weight (weight of a unit cube) of the 
liquid. In this formula, if h is in feet and a is in square feet, w is 
the weight of a cubic foot; hence, for water 

P = 62Aah (2) 

If h is in feet, a in square inches, and w = weight of 1 cu. ft. 

ahw 
144 
and for water, 

P = AVsah = ^ah (4) 

For a curved surface, a must be equal to the projection of the 
surface on a plane perpendicular to the normal through the 
center of gravity. 

Rule. — The total normal pressure upon any submerged surface 
due to the weight of the liquid is equal to the weight of a prism of the 
liquid whose base is equal to the projection of the surface on a plane 
perpendicular to the normal at the center of gravity of the surface 
and whose altitude is the depth of this center of gravity below the 
surface of the liquid. 



P = 



(3) 



28 



ELEMENTS OF PHYSICS 



§4 



50. From the foregoing, it will be evident that the pressure on 
the bottom of a vessel has nothing to do with the shape of the 
vessel that contains the liquid. Thus, referring to Fig. 8, if the 
areas of the bottoms ah of the four vessels shown is the same, and 
the depth of the water is the same in all the vessels, the pressure 
(total pressure) on the bottoms of all the vessels will be the same, 
since their centers of gravity are at the same depth below the 




Fig. 9. 



surface of the liquid. Further, if all the vessels stand on a com- 
mon flat, level surface, and the height of the liquid in each is 
the same, then if all are filled with the same liquid and are con- 
nected together by a pipe, the level of the liquid will still be at the 
same distance above the bases. This fact is strikingly shown in 
Fig. 9, where the vessel B has a much larger cross-section than A. 
If water, for example, be poured into A (or B), it will flow into 
B {A) through the connecting pipe, and when the pouring is 
stopped, it will be found that the water is at the same level in 
both vessels. This phenomenon is expressed in the familiar 




Fig. 10. 

saying "Water seeks its level." If the water level were not the 
same in both vessels, there would be a greater specific pressure on 
the bottom and at the entrance to the pipe in one vessel than in 
the other, and the water would flow from the place of higher pres- 
sure than to that of the lower. 

51. It is to be again emphasized that the total normal pressure 
on any submerged surface does not depend upon the shape of the 



§4 



HYDROSTATICS 



29 



G^9 



vessel that contains the liquid. Referring to (a), Fig. 10, suppose 
dchagef to be a dam, the top and bottom having the shape of a 
rectangle. Then the total normal pressure on the inside face of 
the dam depends only on the projected area of that face and the 
depth of the center of gravity of the face below the surface of the 
Hquid, which is assumed to be level with the top of the dam; and 
it makes no difference how far back the water may extend. It 
also makes no difference what shape of the cross-sec- 
tion of the dam may be, provided the projected 
area is the same; that is, the dam may be 
straight, as in (a), or curved, as in (6). 

52. Combined Pressures. — If the upper sur- 
face of a liquid is free, the pressure at any point 
in the liquid is that due to the depth of the water 
(liquid) at that point; but if the liquid is sub- 
jected to an external pressure, the total specific 
pressure at any point is equal to the sum of the 
specific pressure due to the depth of the point 
below the surface of the liquid and the specific 
pressure due to the external pressure. 

Example. — Referring to Fig. 11, ^ represents a 
cylinder filled with water, the inside dimensions being, 
say, 18 in. diameter and 30 inches long. A pipe B, }4 in. 
in diameter is connected to the cylinder at 6 and is filled 
to the point a with water. If a pressure of 16 pounds 
be applied to the handle C, thus plashing on a 
piston touching the water in the pipe at a, 
what will be the total pressure on the bottom 
of the cylinder? on the top of the cylinder? 
the lateral specific pressure at b? Let the 
vertical distance h between the top of the 
water in the tube and the bottom of the vessel 
be 8 ft., and between b and the bottom of the 
vessel 12 inches. 

Solution. — The specific pressure on the 
bottom of the vessel due to the water in the 
cylinder and pipe is, by formula (2), Art. 48, 
since 8 ft. = 96 in., p = M X 96 = 3.4667 lb. per sq. in. 

The specific pressure on the water at any point in the cylinder due to the 
push on the handle is equal to the pressure on the piston divided by the 

area of the piston, or ^^^^ ^ ^^ = 81.4871b. per sq. in. The total specific 



Fig. 11. 



= 81.487 lb. per sq. in. 
84.954 lb. 



.7854 X .52 
pressure on the bottom is 81.487 + 3.467 = 84.954 lb. per sq. in., and 
total pressure on bottom is .7854 X 18" X 84.954 = 21,618 lb. Ans. 
The specific pressure on the top due to the water in the pipe is p = -g^ 



30 ELEMENTS OF PHYSICS §4 

X (96 — 30) = 2.383 lb. per sq. in., and the total specific pressure on the top 
is 81.487 + 2.383 = 83.87 lb. per sq. in. The total pressure on the top is 
.7854 X 182 X 83.87 = 21,342 lb. Ans. 

The lateral specific pressure at b is 3V(t(96 — 12) = 3.033 lb. per sq. in. 
due to the water only. The total specific pressure at b is 81.487 + 3.033 
= 84.52 lb. per sq. in. Ans. 

Observe that the specific pressure due to the push on the handle is 
transmitted undiminished in all directions within the cylinder, while that 
due only to the water depends on the depth of the point considered below 
the level c. 



EXAMPLES 



(1) Referring to Fig. 2, suppose the weight on top of the piston is 1200 
pounds, the diameter of the piston is 21 in., and the depth of the water under 
the piston is 40 in., what is (a) the total pressure on the bottom of the 
cylinder? (6) on the bottom of the piston? (c) the specific pressure due to 
the weight. f (a) 1700.3 1b. 

Ans. \ (c) 1200 lb. 

I (c) 3.4646— lb. per sq. in. 

(2) Referring to the preceding example, what is the force (total normal 
pressure) tending to separate one half of the cylinder from the other half? 

Ans. 3516.9 lb. 

(3) In Fig. 9, suppose that the diameter of A is 1.72 in. and of B 9.8 in. 
If a piston weighing 12 lb. rests on top of the water in B and a force of 
24 lb. (including weight of small piston) is applied to a piston on top of 
the water in A, what force (weight) must be applied to the piston in B to 
keep it stationary? Ans. 767.12 lb. 

Suggestion. — The weight of the piston in B must be subtracted from the 
upward pressure of the water to find the total downward force required to 
balance the pressure on piston in A. 

(4) Neglecting the weight of the water in the last example, how far will 
the piston in B move when the piston in A moves 3)4 iii-? 

Ans. 0.1 in. very nearly. 

(5) Referring to Fig. 7, suppose abed to be a flat rectangular plate S}4 in. 
by 11 in., and that its center of gravity is 43 in. below the water level; what 
is the total normal pressure on the plate? Ans. 145.18 lb. 

(6) Referring to the example of Art. 52, what will be the total upward 
pressure against the top of the cylinder, if the diameter of the pipe 5 is 2 in., 
the other dimensions and the pressure on the handle C being the same as 
before? Ans. 329,831 lb. 

(7) The total difference of level between the top of the water in a reservoir 
and the nozzle of a fire hose is 225 ft. ; what is the pressure of the water at 
the nozzle when the water is not flowing? Ans. 97.6 lb. per sq. in. 

(8) A certain reservoir has a uniform cross-section shaped like the second 
illustration in Fig. 8. The bottom is a rectangle 144 ft. by 350 ft. ; what is 
the total pressure on the bottom when the depth of the water is 75 ft., 
assuming that the bottom is level? 

Ans. 235,872,000 lb. = 117,936 tons. 



§4 HYDROSTATICS 

BUOYANCY AND SPECIFIC GRAVITY 



31 



BUOYANCY 

53. Conditions under which Bodies Float or Sink. — All bodies 
have a tendency to float; let us consider the reason. Referring 
to Fig. 12, ABCDEFG represents a tank filled with a Hquid in 
which two bodies M and N are immersed. M is a prism, its 
bases being parallel to the upper (flat) surface of the liquid. 
The downward pressure on the body M is equal to the weight of a 
prism of the liquid whose volume is a'h' c' d' abed; the upward 
pressure is the weight of a prism of the liquid whose volume 
is indicated by a'h'c'd'efg; the difference of these volumes is 
the volume of the prism; and the difference between the down- 
ward and upward pressures is equal to the weight of a prism of 




Fig. 12. 

the fluid having the same volume as the prism M. In other 
words, the difference between the downward and upward pressures 
is equal to the weight of the liquid displaced by the body, and this 
statement is true whatever the shape of the body. For instance, 
let N be an irregular body, say a stone; let e'fg' be the projection 
of the stone on the upper surface of the Hquid; then the down- 
ward pressure is equal to the weight of a column of the liquid 
having the shape e'f'g'edg; the upward pressure is the weight 
of a column of the liquid having the shape e'f'g'efg; and the dif- 
ference between these pressures is the weight of a body of fluid 
having the same shape as the stone. Further, it makes no dif- 
ference how far below the surface of the liquid the body lies, 
the weight of a body of liquid having the shape of the submerged 
body will be the pressure tending to force the body upward. 



32 ELEMENTS OF PHYSICS §4 

Since gravity tends to pull the body down and the difference 
between the downward and upward pressures tends to push it 
up, the force urging the body down through the liquid is equal to 
the weight of the body minus the weight of an equal volume of 
the liquid. If, therefore, the density of the body is the same as 
that of the liquid, the body will stay in any position in the liquid 
and at any depth that it may be placed, assuming the density of 
the liquid to be uniform. If the density of the body is less than 
that of the liquid, the body will float, that is, only a part of it 
will be submerged, the weight of a volume of liquid equal to the 
submerged part being equal to the weight of the body. If the 
density of the body is greater than that of the submerged part, 
the body will sink, that is, it will fall through the liquid until it 
touches bottom. 

As an illustration of the statements in the last paragraph, place 
an egg in a can of water; the egg will sink, showing that its 
density is greater than water. Now, leaving the egg in the water, 
dissolve in it some salt (or sugar), stirring the water so it will 
be equally dense throughout. The salt being denser than the 
water, the density of the water will gradually increase until it 
becomes the same as that of the egg, and when this point is 
reached, the egg will stay in any position in the water and at 
any depth. As more salt is added and dissolved, the water 
becomes denser than the egg, and the egg will rise and float, a 
part of it extending out of the water. 

54. The tendency of any body to float when immersed in a 
fluid is called buoyancy, and the denser the body the less buoyant 
it is. What is true in this respect of a liquid is equally true of a 
gas, the upward or buoyant force being equal to the weight of a 
body of gas having the same volume as the submerged body. 
In the case of a balloon, the gas (hydrogen) with which it is filled 
is very much lighter than air, a cubic foot of air weighing about 
14.5 times as much as hydrogen. Assuming that a cubic foot 
of air weighs .08 pound under certain conditions, 100,000 cubic 
feet will weigh 8000 pounds, and 100,000 cubic feet of hydrogen 
under the same conditions will weigh 8000 -^ 14.5 = 552 — pounds. 
The difference is 8000 — 552 = 7448 pounds, which is the buoyant 
effect, or force urging the balloon upward. The total upward force 
is 8000 pounds, but this is counteracted by the weight of the 
hydrogen, the weight of the balloon, and the load lifted. As the 
balloon moves up, the air becomes less and less dense, and a point 



§4 HYDROSTATICS 33 

will be reached where the balloon can ascend no higher, remaining 
stationary (in still air) unless some of the hydrogen be allowed to 
escape, when the balloon will descend ; or, if the balloon carries 
ballast and this be thrown out, the balloon will rise higher. 

Mercury, which is a liquid at ordinary temperatures, is so 
dense that iron, copper, lead, etc. will float in it; but gold, 
platinum, tungsten, etc. will sink in it, because their densities 
are greater than the density of mercury. 

55. Weight in a Vacuum. — A vacuum is a closed space from 
which all air or other gas has been removed; the inside of a 
carbon filament lamp is a good example of a vacuum. The air 
must be removed, since it would otherwise unite with the carbon 
(when heated) and the filament would be destroyed. A little 
consideration will show that a body will weigh more in a vacuum 
than in air, because the buoyant effect of the air counteracts 
by just that much the force of gravity. For example, if a cubic 
foot of water weighs 62.4 pounds in air when the air weighs 
.08 lb. per cu. ft., a cubic foot of water will weigh 62.4 + -08 
= 62.48 pounds in a vacuum. The weight in a vacuum is there- 
fore the true weight of a body. However, in all practical and 
commercial transactions, the weight of a body in air is the weight 
that is used; it is only in accurate scientific calculations that 
weights are expressed as the true weights in a vacuum. 



SPECIFIC GRAVITY 

56. Definition. — The specific gravity of a body is the ratio of 
the weight of the body to the weight of an equal volume of water. 
Thus, the weight of a cubic foot of cast iron is commonly taken 
as 450 pounds, and (when finding the specific gravity) the weight 
of a cubic foot of water is 62,4 pounds; hence, the specific 

450 
gravity of cast iron is w^ = 7.21. In other words, cast iron 

weighs 7.21 times as much as an equal volume of water at its 
temperature of maximum density. When 62.4 pounds is used 
as the weight of a cubic foot of water in finding the specific 
gravity, it is useless to find the value of the ratio to more than 
three significant figures. 

62.4 

The specific gravity of water is evidently 1, since ^j = 1. 

Consequently, if the specific gravity of a body is greater than 1, 



34 ELEMENTS OF PHYSICS §4 

the density of the body is greater than the density of water, and 
the body will sink in water; but if the specific gravity is less than 
1, the density of the body is less than that of water, and the body 
will float in water. 

57. Let s = the specific gravity of a body, W = its weight, 
and w = weight of an equal volume of water; then, 

s = — (1) 

w 

W = ws (2) 

The specific gravity of most materials and substances can be 
obtained from printed tables; hence, if the volume and specific 
gravity of any body or substance is known, its weight can read- 
ily be found. For, the weight of an equal volume of water is 
62.4 V, where V is the volume in cubic feet, and this equals w 
in formula (2) ; therefore, substituting in formula (2) , 

W = 62.4 Vs (3) 

If y = the volume in cubic inches, 1 cu. in. of water weighs 
s\% pounds (see Art. 48), and 

TT = ^ (4) 

360 ^ -^ 

Example. — Taking the specific gravity of silver as 10.53, what is the 

weight of 8.65 cu. in.? 

Solution. — Applying formula (4), v = 8.65 and s = 10.53; hence, 

„ 13 X 8.65 X 10.53 „ ^^^ ,, . 

W = kt;^ = 3.289 lb. Ans. 

360 

58. When using the metric system of weights and measures 
in connection with specific gravity problems, the cubic decimeter 
is taken as the standard of volume. Since 1 cubic decimeter 
= 1 liter, and since, by definition, 1 kilogram is the weight of 1 
liter of water, then letting W = the weight of one cubic decimeter 

W 

of the substance, formula (1) of the last article becomes s = -3- 

= W; that is, the specific gravity of any substance is numerically 

equal to the weight in kilograms of 1 cubic decimeter of the 

substance. As an illustration, the weight of 1 cubic foot of 

cast iron is 450 pounds; since 1 kg. = 2.2046 pounds and 1 cu. 

A A1 no^ -1 A f ,■ • u 450X61.024 

dm. = 6L024 cu. m., 1 cu. dm. of cast iron weighs .„ncy w o r>r^An 

1728 X 2.2046 

= 7.21+ kg. The value previously found for the specific 

gravity of cast iron was 7.21; hence, the numerical values 7.21 

agree. 



§4 HYDROSTATICS 35 

To find the specific gravity of a liquid, all that is necessary is 
to fill a hter measure, weigh it, and then subtract the weight of 
the measuring vessel; the result expressed in kilograms will be 
the numerical value of the specific gravity of the hquid. 

To find the specific gravity of a sohd, it would be quite dif- 
ficult to form it into the shape of a cube measuring 1 cu. ft. or 1 
cu. dm. on each edge ; in fact, in many cases it would be impossible. 
Consequently, the usual method is to immerse the object in 
water and note the loss in weight. This loss in weight is evidently 
equal to the weight of a volume of water equal to the volume of 
the sohd, the value of W in formula (1), Art. 67. Let W 
= the weight in air and W = the weight in water; then 

^ W - W 
For example, suppose a certain small object weighs 367 grains 
in air and 310 grains in water; its specific gravity is s = 
367 

= 6.44-. 



367 - 310 

59. Distinction between Density and Specific Gravity. — 

Density may be defined as specific mass or as specific weight, 
according to which of the two definitions of Art. 30 is used. 
The word specific, as used in physics, always impHes a reference 
to some standard. Thus, specific weight means the weight of a 
unit of volume (1 cu. ft., 1 cu. dm., etc.), specific pressure is the 
pressure per unit of area (1 sq. ft., 1 sq. cm., etc.), specific volume 
is the volume of a unit of weight (1 lb., 1 kg., etc.). If, therefore, 
density be defined as the weight of a unit of volume and the unit 
of volume be taken as 1 cu. dm. = 1 liter, the numerical values 
of the density and specific gravity of any body will be equal; in 
fact, many writers on scientific subjects use these two terms 
interchangeably. They do not, however, mean exactly the same 
thing, since density means specific weight (or specific mass), 
while specific gravity means the number of times heavier a 
substance is than an equal volume of water. To prevent any 
ambiguity, the first definition of density, according to which 

^ ^ V ^ gV' ^^^^ ^^ "^^^ hereafter, except for gases. 

60. Specific Gravity of Gases.— Gases are so much lighter 
than equal volumes of Hquids and sohds that is customary to 
express their specific gravities as the ratio of the weight of any 
volume of gas to the weight of an equal volume of air. The 



36 



ELEMENTS OF PHYSICS 



§4 



weight of any gas depends not only on its volume but also on its 
temperature and pressure. The weight of a cubic foot of air at 
32° F. (0° C.) when its pressure is 14.695 pounds per sq. in. is 
.08071 pound. (This combination of pressure and temperature 
is called standard conditions.) Hence, if the specific gravity 
of nitrogen be given in a certain table as .970, the weight of a cubic 
foot at 32° F. and a pressure of 14.695 lb. per sq. in. is .08071 
X .970 = .07829 pound. Properties of gases will be considered 
later. 



HYDROMETERS 

61. Definition. — A hydrometer is an instrument by means of 
which the specific gravity of a hquid may 
be found; they are made in many forms 
and are called by various names, but they 
all depend for their action on the fact that 
if a body lighter than the liquid in which it 
is placed sink to a certain mark on the 
body, it will sink farther in a lighter liquid 
(one less dense) and not so far in a heavier 
liquid (one of greater density). Fig. 13 
shows two forms of the instrument, the 
one at (a) being for liquids lighter than 
water and the one at (b) for liquids heavier 
than water. As will be seen, they consist 
of glass tubes closed at both ends and con- 
taining a graduated scale. One end is 
loaded with mercury or shot to make the 
instrument stand upright when placed in 
the liquid. The enlarged part above the 
loaded end increases the buoyancy. The 
instrument shown at (a) has a narrow stem, 
increasing its sensitiveness, and adapting 
it to liquids heavier than water; the one 
shown at (6) has a wide stem, and is for 
use in liquids lighter than water. 

The density of water and other liquids 
usually decreases as the temperature in- 
creases; for this reason it is necessary to 
have a standard temperature for the gradu- 
ation of the hydrometer and for the liquid when the hydro- 



FiQ. 13. 



§4 HYDROSTATICS • 37 

meter is placed in it: this temperature is generally 15° C. or 60° F. 
C. and F. mean centigrade and Fahrenheit respectively, and will 
be explained later. Although 15° C. really equals 59° F., the 
difference between 59° and 60° is so slight that it may usually 
be neglected. 

62. The Graduations on the Scales. — The main graduations 
on any hydrometer scale are called degrees, and these are fre- 
quently subdivided into tenths of a degree. Some instruments are 
so divided that they give the specific gravity while in others, if 
the specific gravity be desired, it must be calculated from the scale 
reading. The scale most commonly used in the United States and 
Canada is the Beaume; and while there are several formulas for 
calculating the Beaume scale, the so-called American Beaume 
is the only one that will be considered here. Any reference to the 
Beaume scale in this course will be understood to be the American 
Beaume at 60° F. 

Let s = the specific gravity of the Uquid to be tested at 60° F., 
and let 5 = the reading of the scale in degrees Beaume; then 
for liquids lighter than water, 

_ 14 .^, 

' ~ ISO + B ^^^ 

For liquids heavier than water. 

For hquids lighter than water, the scale evidently begins at 
10, since by formula (1) s = ^^^ ^^ =1 = the specific 

gravity of water. And for liquids heavier than water, the scale 

145 
begms at 0, since by formula (2), s = ^^ ^ = 1. 

To use the instrument, partly fill a glass container, of such 
depth that the hydrometer will not touch bottom, with the Hquid 
to be tested, first being sure that its temperature is 60°, then 
gently insert the hydrometer, and when it comes to rest in an 
upright position, note where the upper surface of the hquid 
crosses the scale; this will be the reading in degrees Beaume. If 
the specific gravity is desired (which is not usually the case, the 
reading in degrees Beaume being generally sufficient), calculate 
it by substituting the reading for B in formula (1) or (2). 

If it be desired to convert specific gravity into degrees Beaum^, 



38 ELEMENTS OF PHYSICS §4 

simply solve the above formulas for B, obtaining for liquids 
lighter than water, 

B = ]^-^ (3) 

and for liquids heavier than water, 

B = iM(i^i) (4) 

s 
Example 1. — Suppose a certain liquid that is known to be heavier than 
water has a density of 18.4° Beaume; what is its specific gravity? 
Solution. — Substituting 18.4 for B in formula (2), 

s = -TTs r5-ir = 1.1453. Ans. 

145 — 18.4 

Example 2. — -Knowing that the specific gravity of a certain liquid is .886, 

what is the density in degrees Beaume? 

Solution. — Since the specific gravity is less than 1, use formula (3), and 

„ 140 - 130 X .886 _^o-R , , 

t> = t^ttt; = 28 Beaume. Ans. 

.886 

63. Object of Hydrometer. — The reason for using the hydrom- 
eter to get the comparative densities of liquids is that materials 
bought and sold in liquid form (mixtures of several different 
materials) do not contain a constant percentage of some particu- 
lar substance for which the liquid was bought. As a simple 
illustration, consider a mixture of common salt and water; the 
greater the percentage of salt in the solution the greater the den- 
sity of the solution. In a solution of salt and water, if the den- 
sity of the solution or its specific gravity be known, then by means 
of a special formula or table, the percentage of salt can be found. 
This same method can be used to find the percentage of other 
substances in solution. In many pulp and paper mills, the 
hydrometer is used for taking the strength of the bleach liquor; 
if the bleach is of poor quality, that is if it contains soluble im- 
purities, as calcium chloride, which remains over when the bleach 
powder decomposes, misleading results will be obtained with the 
hydrometer, because the calcium chloride increases the density 
just as the bleach itself does. Consequently, before dependence 
can be placed on the results obtained with the aid of the hy- 
drometer, one must be sure that the solution itself is pure. 



EXAMPLES 

(1) A liter of hydrogen under standard conditions weighs .089873 gram, 
a liter of air weighs 1.2929 gram ; (a) what is the specific gravity of hydrogen ? 
how many liters of hydrogen equal the weight of 1 liter of air? 

. / (a) .069514 
^""^•XQ}) 14.389 



§4 



HYDROSTATICS 



39 



(2) A flask holding 50 c.c. (cubic centimeters) is filled with a liquid and 
weighed; after deducting the weight of the flask, the weight of the liquid is 
found to be 64.6 g. (grams); (a) what is the specific gravity of the liquid? 
(6) what is its relative density in degrees Beaume? 

, ( (a) 1.292 

^'''- I (6) 32.77» Be. 

(3) What IS the weight of a cold drawn steel rod 2}4 in. in diameter and 
13 ft. long, the specific gravity of the steel being 7.83? Ans. 175.4 lb. 

(4) The specific gravity of oxygen is 1.1053; what is the weight of 468 
cubic feet under standard conditions? Ans. 37.77 lb. 

(5) Suppose a liquid that is lighter than water has a relative density of 
42° Beaume; what is its specific gravity? Ans. .814. 

(6) The relative density of a certain liquid that is lighter than water is 
17.4°Beaume; what is the weight of 50 c.c? Ans. 47.5 g. 



CAPILLARITY 

64. Capillary Attraction. — The word capillary means hair-like, 
and as used in physics, it refers to small or fine tube-like holes and 
channels. If a clean glass rod be placed in a vessel of water, as 
shown in (a), Fig. 14, it will be found that the surface of the 
water around the rod is not level, but curved, the water being 
drawn up about the rod, and the 
part touching the rod will be higher 
than the water level as at a and b. 
If, instead of a glass rod, a glass 
tube be used, the water in the tube 
will be higher than the water level 
outside (as at c in Fig. 16); the 
shape of the upper surface of the 
water in the tube will also be 
curved, the outer edge being higher 
than the center. It will also be found that the smaller the 
diameter of the hole the higher the water will rise in the tube. 
This phenomenon is called capillary attraction, and the results 
noted are caused by the adhesion of the water to the tube and 
the cohesion of the water particles. Fibers used in making paper 
are really fine tubes and consequently behave similarly. 

If the glass tube be covered inside and out with a thin coating 
of grease, an exactly opposite effect will be obtained; the water 
will be depressed inside and outside of the tube, as shown at (6), 
Fig. 14, and the curves will be reversed. This latter effect will 
also be observed if a clean glass tube be inserted in mercury. 




(b) 



Fig. 14. 



40 



ELEMENTS OF PHYSICS 



§4 



It will be found on examination, that in the first case, the liquid 
wet the tube and in the second case, it did not wet the tube. In 
general, capillary attraction draws the Hquid up when the liquid 
wets the surface and pushes it down when the liquid does not wet 
the surface. A striking illustration may be obtained by means of 
two flat panes of glass, partly immersed in a vessel of water, as 
shown in Fig. 15. Bringing two of the ends together and sep- 
arating the other two ends slightly, as shown, the water will be 
found to rise in a curve, the highest point being at the ends that 
touch. 




Fig. 15. 



If one end of a lump of sugar be touched to and held in contact 
with the surface of a liquid, as water, milk, etc., the entire lump 
will soon become wet, the liquid being raised through the pores by 
capillary attraction. When a candle burns, the wax or tallow is 
melted (changed to a liquid) and capillary attraction draws the 
melted wax or tallow up through the vnck to the flame. If there 
were no wick, the candle would not burn in the regular manner. 

65. The vertical height through which a liquid will be raised by 
capillary attraction above the true level of the liquid varies in- 
versely as the diameter of the tube, for the same liquid. Thus, 
water will rise higher than alcohol, but for the same liquid in two 
tubes A and B, if A is twice as large as B, the liquid will rise only 
one-half as far in A as in B; if the diameter of A is .03 in. and of B 
.01 in., the liquid will rise .03 -t- .01 = 3 times as far in B as in A. 



§4 



HYDROSTATICS 



41 



Capillary attraction need be considered only when the diameter 
of the tube is quite small. The curve formed by the top of the 
liquid is called the meniscus, and it is crescent or bow-shaped. 
The meniscus at c, Fig. 16, is said to be concave upward, and that 
in (6), Fig. 14, is said to be convex upward. The top of the 
meniscus, in either case, is the point where the axis of the tube 
intersects it. When reading the height of a column of fluid that 
has a meniscus, it is usual to take the top of the meniscus as the 
top of the column. This applies to thermometers, barometers, 
pipettes, burettes, etc. 

An interesting example of capillary attraction is shown in Fig. 
16. Here A and B are glass vessels connected at D. C is a small 
glass tube also connected to D. 
The diameters of A and B are such 
that the water level in each is the 
same, and is indicated by the line 
ah. The water level in C, how- 
ever, is much higher, being indi- 
cated by c. The distance cd 
represents the vertical height that 
the water was raised by capillary 
attraction, and it does not in any 
way increase the hydrostatic pres- 
sure on the water in A and B, and it would evidently not be 
correct to measure the height of c in any calculation regarding 
the normal pressure at any point of the liquid in the vessels. 

66. Some Examples of Capillary Action. — Capillary action 
plays a very important part in ordinary everyday affairs. With- 
out capillary action, no plant could live, since capillary attraction 
is what draws the sap up from the ground. I^ is capillary attrac- 
tion that enables a piece of blotting paper to absorb ink, that 
brings oil to the flame through the wick of a lamp, that causes 
wood to swell when placed in water. In coloring or sizing paper, 
capillary attraction helps by drawing solutions into the fibers. 
It can be made to exert an immense force. Thus, if a hemp rope 
be drawn tight and is then wet with water, it contracts, because 
the fibers run around in helixes (somewhat like a screw thread) ; 
as the capillary attraction draws up the water, the rope swells and 
the fibers shorten in a direction lengthwise of the rope the re- 
sult being a tremendous pull. 




Fig. 16. 



42 ELEMENTS OF PHYSICS §4 



PNEUMATICS 



THERMOMETERS 

67. Pneumatics is that branch of science that treats of the 
properties of air and gases; it might be called the hydrostatics of 
gases. 

As was previously pointed out, all gases completely fill the 
vessels that contain them; it is also to be noted, that the density 
of a gas is uniform through its entire extent, that is, no matter 
what the shape of the container, a cubic inch, say, of the gas will 
have the same density regardless of what part of the vessel it is 
taken from. Theoretically, this last statement is not quite true, 
since a cubic inch taken from the top of a vessel will weigh a 
trifle less than that taken from the bottom, the column of gas 
having weight in the same manner that a column of water has 
weight; but, in practice, this may be neglected, and the density of 
a gas may be considered to be uniform throughout the container. 

68. Tension. — The molecules of gases are in rapid vibration, 
and they are constantly trying to escape from the vessel that con- 
tains the gas; the result is that the gas exerts a pressure against 
the walls of the container in much the same manner that a 
helical spring exerts a pressure when a load is placed on it. The 
condition of the gas or spring that causes it to exert pressure is 
called tension. The force with which the spring presses against 
the load is equal to and measures its tension, and the pressure 
which gases exert against the walls of their containers also equals 
and measures their tension. Tension is measured in the same 
manner and in the same units as pressure, and the word pressure 
is frequently used instead of tension, since the tension must 
always equal the pressure. 

69. Before the tension of any gas can be found, it is necessary, 
as will be more fully explained later, to know its temperature, 
because any increase or decrease in the temperature increases 
or decreases the tension, provided the volume of the gas does not 
change. Under ordinary conditions, temperatures of bodies 
are measured by instruments called thermometers. 

70. Thermometer Scale. — All thermometers depend for their 
operation upon the fact that most substances expand when heated 



§4 PNEUMATICS 43 

and contract when cooled; if this variation in length, area, or 
volume can be measured, a thermometer can be constructed. 
The substance most generally used for this purpose is mercury. 
As usually constructed, a glass tube with a bulb at one end is 
partly filled with mercury; the air is then exhausted above the 
mercury and the other end of the tube is sealed. When the tube 
is heated, the mercury expands, and the column in the tube 
lengthens; and when the tube is cooled, the mercury contracts, 
and the column shortens. The hole in the tube is quite fine, so 
that a small expansion in the volume of the bulb will make a 
considerable increase in the length of the column. 

In constructing a scale for a thermometer, it is assumed that the 
ice made from pure water always melts at the same temperature 
and that pure water always boils at the same temperature, 
when the pressure conditions are the same. The tube is therefore 
inserted in melting ice, and the height of the mercury column 
is marked on the scale. It is next inserted in boihng water, 
and the height of the mercury column is marked on the scale. 
It is now assumed that if the distance between these two points 
be divided into equal parts, each part will indicate an equal rise 
(or fall) in temperature. Experience has shown these assumptions 
to be correct. 

There are two thermometer scales in general use— the Fahren- 
heit scale and the centigrade or Celsius scale. The former is 
in common use in English speaking countries, and the latter in 
all other countries (except Russia) and by scientists generally. 
In both scales, the principle divisions are called degrees, and 
these are subdivided into tenths, hundredths, etc., the degrees 
on the Fahrenheit scale being denoted by N°F. and those on 
the centigrade scale by N°C., N representing the number of 
degrees; thus, 59 °F. and 15°C., represent 59 degrees Fahrenheit 
and 15 degrees centigrade, respectively. 

71. When graduating a thermometer (thermometer means 
heat measurer) in accordance with the Fahrenheit scale, the 
point that indicates the temperature of melting ice is marked 
32°, and the point that indicates the temperature of boiling 
water is marked 212°; the difference is 212° - 32° = 180°, 
and the distancq between these two points is divided into 180 
equal parts, each of which is 1°F. These divisions are carried 
above and below the two fixed points, those below being numbered 
29, 28, etc. until is reached, which will be 32° below the melting 



44 ELEMENTS OF PHYSICS §4 

point of ice. All graduations below are negative, and they 
increase numerically from 1 up ; thus — 12°F. means 12 degrees 
below zero on the Fahrenheit scale. 

The centigrade scale (sometimes called the Celsius scale) is 
graduated according to the same plan. The point indicating the 
temperature of melting ice is marked 0°; the point indicating 
the temperature of boiling water is marked 100°; and the distance 
between them is divided into 100 equal parts. Since this distance 
is the same on both scales, 1°C. : 1°F. = 180 : 100 from which 
it is seen that 1°C. = \U = W°; and 1°F. = {U = |C°. 

72. To convert degrees C. into degrees F., multiply the reading 

in degrees C. by f, and the result will be the number of F° 

above the temperature of melting ice; to find the temperature 

above 0°F., add 32° to the product. Expressed as a formula 

F.° = C.°X 1 + 32 (1) 

To convert degrees F. into degrees C, subtract 32° from the 
reading and multiply the remainder by I; the product will be the 
degrees C. Expressed as a formula. 

C°. = (F°. - 32°) Xf (2) 

Example 1. — How many degrees F. are equivalent (a) to 15°C.? (6) 
to -20°C.? 

Solution.— (a) Applying formula (1), 15 X I + 32 = 59°F. Ans. 

(b) Applying formula (1), -20 X f + 32 = -36 + 32 = -4°F. Ans. 

Example 2. — (a) How many degrees C. are equal to 240°F.? (b) to 
- 22°F? 

Solution.— (a) Applying formula (2), (240 - 32) X f = 115f°C. Ans. 
(6) Applying formula (2), (-22 - 32) X f = - 30°C. Ans. 

In both of the above examples, the numbers given are supposed to be 
thermometer readings. 



THE ATMOSPHERE 

73. Perfect and Imperfect Gases. — A perfect gas is one that 
remains a gas under all conditions of temperature and pressure; 
such a gas is also called a permanent gas. There is really no 
such thing as a perfect gas, since all gases have been liquefied 
at extremely low temperatures and under very high pressures. 
For practical purposes, however, hydrogen, nitrogen, oxygen, 
atmospheric air, etc. may be considered to be perfect gases. 
An imperfect gas or vapor is one that is readily liquefied under 
ordinary conditions; thus, steam, which is a vapor of water 
(and a gas) condenses when allowed to expand, and is, therefore, 



§4 



PNEUMATICS 



45 



an imperfect gas. In what follows, air will be taken as an example 
of a perfect gas, and the laws relating to air will apply to all 
other perfect gases. 

74. The Atmosphere. — The earth is entirely surrounded by an 
invisible gaseous envelope called the atmosphere, and the 
mixture of gases composing the atmosphere is called air. The 
principal gases composing the air are nitrogen and oxygen; 
in addition to these, there is also a small percentage of carbon 
dioxide, water vapor, and minute quantities of several other gases. 
For practical purposes, the thickness of the gaseous envelope 
(depth of the atmosphere) is about 50 miles, but the actual depth 
may be several hundred miles. The density is greatest at the 
surface of the earth, but greater, of course, at the bottom of a deep 
well or shaft, and decreases as the distance above sea level 
increases. 

75. The atmosphere exerts a pressure due to its weight on 
everything it touches; that this is the case can be shown in 
many ways, but most strikingly in the following manner: Take 
a glass tube that is about 40 inches 
long and closed at one end. Fill the 
tube with mercury, place the hand 
over the open end, and keeping it 
there, invert the tube; the mercury 
will be felt pressing against the hand. 
Still keeping the end of the tube 
covered, insert the hand^ and end of 
tube in a dish of mercury, and re- 
move the hand. If the tube is verti- 
cal, it will be found that the distance 
between the top of the column and 
the top of the mercury in the dish is 
about 30 inches. See Fig. 17. This 
column of mercury is supported by the pressure of the atmos- 
phere; because, the weight of the mercury exerts a pressure on 
the mercury in the dish that is transmitted in all directions, as 
in the case of any other fluid, and the upward pressure would 
raise the level of the liquid (mercury) in the dish were it not for 
the pressure of the atmosphere that just counterbalances it. 

76. The space between the top of the mercury column and the 
closed end of the tube, which is the difference between the length 




Fig. 17. 



46 ELEMENTS OF PHYSICS §4 

of the tube above the level of the mercury in the dish and the 
length (height) of the mercury column, is entirely empty, or in 
other words, a vacuum; this is called a Torricellian vacuum, 
after Torricelli (1608 — 1647), who first performed this experiment 
and first showed that the air had weight. If the pressure on the 
surface of the mercury in the dish be increased, the column of 
mercury in the tube will rise, and if the pressure be decreased, 
it will fall; this shows conclusively, that the column is supported 
by the pressure on top of the mercury in the dish. If the tube be 
inclined, as shown in Fig. 17, so that the vertical distance be- 
tween the top of the tube and the level of the mercury in the dish is 
less, the mercury rises in the tube until it fills it when this vertical 
distance equals about 30 inches. The top of the mercury in AB 
and the top in the tube B' are then in the same horizontal line. 

77. The specific gravity of mercury is 13.6 at 32°F., and the 
weight of 1 cu. in, = .49111 lb.; hence, a column of mercury 
1 in. high exerts a pressure of .49111 lb. per sq. in. The 
height of the mercury column due to the atmospheric pressure 
varies considerably, being dependent on the altitude of the place 
above sea level, the temperature, the amount of water vapor 
contained in the air, etc., but the standard value is 760 milli- 
meters at 0°C., at sea level; this corresponds to 29.9213 inches 
at 32°F. (= 0°C), at sea level. The pressure exerted by the 
atmosphere is therefore 29.9213 X .49111 = 14.6946 lb. per sq. 
in. = 14.6946 X 144 = 2116 lb. per sq. ft. This pressure is 
called 1 atmosphere, and whenever pressures are given in atmos- 
pheres, they can be changed into pounds per square inch by multi- 
plying the number of atmospheres by 14.6946, which is usually 
expressed as 14.7. Since 1 lb. per sq. in. = 703.09456 Kg. per 
sq. m., 1 atmosphere is equal to 703.09456 X 14.6946 = 10,332 
kilograms per square meter. Since the specific pressure of 1 
atmosphere is 2116 lb. per sq. ft., this means that everything on 
the earth's surface is subjected to a pressure of over a ton on 
every square foot. The reason that we do not notice this 
enormous pressure is that the air within the body has practi- 
cally the same tension as the pressure of the air outside, and 
one counteracts the other. It becomes very apparent, however, 
on top of a high mountain or when in a balloon or airplane at 
a great height from the earth; the tension of the air within the 
body is then greater than the pressure outside, and may result 
in the bursting of small blood vessels. 



§4 PNEUMATICS 47 

78. Partial Vacuum. — A perfect vacuum is a closed space 
that contains nothing that can exert a pressure on the walls that 
enclose the space. For practical purposes, the Torricellian 
vacuum mentioned in Art. 76 is a perfect vacuum; it is not ac- 
tually so, because there is a very small amount of air present, 
probably held in suspension within the mercury, but the pressure 
it exerts is so small that it can hardly be detected. The space 
within a carbon-filament electric-light bulb is very nearly a per- 
fect vacuum. 

If, in Fig. 17, a connection be made between the top of tube B 
and a receptacle containing air (or other gas), and a little air be 
admitted above the mercury column, the pressure of this air will 
counterbalance a like weight of mercury; the column will then 
shorten by this amount. Thus, suppose the pressure of the air 
enclosed above the mercury is 2 lb. per sq. in.; this is equiva- 
lent to 2 4- .49111 = 4.0724 inches of mercury, and the height of 
the mercury column will then be 29.9213 - 4.0724 - 25.8489 
in., say 25.85 in. The space above the mercury column is then 
called a partial vacuum. Partial vacuums are nearly always 
measured by the number of inches of mercury that will be sus- 
tained by the difference between the pressure within the partial 
vacuum and the pressure of the atmosphere. For instance, the 
partial vacuum just referred to would be called a vacuum of 
25.85 inches. It may be remarked that vacuum gauges are 
graduated in inches of mercury and read from to 30 inches. 

79. For rough calculations, the weight of a cubic inch of mer- 
cury is taken as }y^ pound and the pressure of the atmosphere as 
15 pounds per square inch; consequently, a vacuum of, say, 19 
in. represents (roughly) a mercury column of 19 X .5 = 19 ^ 2 
= 9.5 lb., and the pressure within the partial vacuum is 15 — 9.5 
= 5.5 lb. per sq. in. 

If exact results are desired, use the following formula in which 
m = inches of vacuum and p = pressure in pounds per square 
inch within the partial vacuum: 

p = 14.6946 - .491 llw 

Applying this formula to the preceding case, p = 14.6946 
- .49111 X 19 = 5.3635 lb. per sq. in. 

80. Examples of partial vacuums are very numerous. When 
soda water is sucked through a straw, the air in the mouth is 
drawn into the lungs, a partial vacuum is created in the mouth, 



48 



ELEMENTS OF PHYSICS 



§4 



and the pressure of the atmosphere forces the jfluid up through the 
straw. The common suction pump, the siphon, etc. all operate 
by reason of a partial vacuum. What are called vacuum pumps 
are used to exhaust air from the suction boxes underneath the 
wire of a paper machine, with the result that as the pulp on the 
wire passes, the water is forced through by the fact that the 
atmospheric pressure above it is greater than the pressure in the 
partial vacuum below it, and the water is said to be sucked 
through the wire. 

81. Action of a Suction Pump. — The manner in which a suction 
pump acts will be clear after considering the diagrammatic 





(«>>. 



(6) 



Fig. 18. 



sections in Fig. 18. D is the pump barrel in which works a 
piston P, the piston being forced up and down by pushing and 
pulling on the piston rod R. The pipe A connects the pump 
barrel to the water supply, and where it joins the barrel is a valve 
V, which is hfted as shown when the water is entering the pump. 
The piston also has valves v' and v". When the piston is moving 
upward, as shown at (a), it leaves a partial vacuum in the space 



§4 PNEUMATICS 49 

B below it; the atmospheric pressure on top of the open water 
causes the water to rise and fill this partial vacuum and follow the 
piston upward. When the piston stops, the water in B tends to 
fall downward, but the weight of the water acts as a pressure on 
top of the valve v and closes it, thus keeping the water in B. 
When the piston moves down, as shown in (6), valves v' and v" 
lift, the water in B passes through the openings in the piston, 
permitting the piston to move down. When the piston again 
moves upward, the weight of the water above it closes the valves 
v' and v", and the water in B' is lifted with the piston and dis- 
charged through C; at the same time, a partial vacuum is created 
in B, the atmospheric pressure forces the water up the pipe A, 
lifts valve v, and follows the piston, as before. 

82. Height of Lift of a Suction Pump. — Theoretically, the 
maximum height of "suction" is equal to the height of a column 
of water that will just balance a perfect vacuum. In the case of 
mercury, this is 29.9213 in., and as the specific gravity of mercury 
is 13.6, the height of a water that will produce the same specific 
pressure is 29.9213 X 13.6 = 406.93 in. = 33.91 ft. Conse- 
quently, if the piston left a perfect vacuum behind it, and there 
were no friction or other resistances to the movement of the water, 
a suction pump would lift water about 34 ft. In practice, from 
28 to 30 ft. is the greatest height that water can be raised with a 
suction pump. It might be thought that since the specific weight 
of hot water is considerably less than that of cold water, hot 
water could be raised to a greater height than cold water; such, 
however, is not the case, the reason being that hot water gives 
ofl" vapor (steam), and the tension of this vapor creates a pres- 
sure that tends to destroy the vacuum. In fact, hot water can- 
not be raised to as great a height as cold water; the hotter the 
water the shorter the lift. 

83. Work Required to Operate a Pump. — Since work is equal 
to force (pressure) multiplied by the distance through which it 
acts, the work required to operate a pump is equal to the distance 
moved by the piston in one stroke multiplied by the lifting force 
applied to the piston and this product multiplied by the number 
of lifting strokes. A stroke is the distance passed through by the 
piston between its lowest and highest positions. The force 
required to lift the piston is the weight of a column of water 
having the same diameter as the piston and whose height is equal 



50 ELEMENTS OF PHYSICS §4 

to the difference of level between the point of discharge and the 
level of the water in the reservoir or other source of supply, indi- 
cated by h in (a), Fig. 18. That this is true is readily seen. 
Thus, when the piston moves up, it lifts a column of water 
above it equal to the weight of a column of water having a 
diameter equal to that of the piston and a length equal to the 
stroke; at the same time the piston creates a partial vacuum that 
causes the water to follow the piston, and the amount of water 
that thus follows the piston is equal to the amount discharged in 
one stroke. The effect is exactly the same as though the water 
above the piston had been lifted the entire distance h. Letting 
h = height of lift in feet, d = diameter of piston in inches, 
s = stroke in inches, the volume of water lifted per stroke is , 7854rf^s 
cu. in. and the weight is .7854:dhX J/o = .028362d2s pounds; 
this multiplied by h in feet is the work done during one lifting 
stroke. Letting w = work in foot-pounds for one lifting stroke, 

w = .028SQ2d^sh 

Example. — Suppose that when the piston is at the upper end of its stroke, 
the distance between the bottom of the piston and the level of water supply- 
is 23 ft. and that the distance between the bottom of the piston and the 
point of discharge is 57 ft. If the diameter of the piston is 8 in. and the 
stroke is 12 inches, what is the work done during one lifting stroke? 

Solution. — The total lift of the water is 23 + 57 = 80 ft. = h; the water 
on top of the piston must be raised 12 in., the length of the stroke; that is, a 
column of water 57 ft. high must be raised 12 in. The height of suction is 
23 ft. and the amount of water discharged during one stroke must be raised 
through this height also. Consequently, the water discharged must be 
raised through a total height of 57 + 23 = 80 ft. Substituting in the form- 
ula the values given, 

w = .028362 X 82 X 12 X 80 = 1742.6 - foot-pounds. Ans. 

The actual work done would be considerably greater on account of fric- 
tion and other resistances. 

84. The Siphon. — A siphon is essentially a bent pipe or tube 
that conveys a liquid from a point of higher level to one of lower 
level, the highest point of the pipe being higher than the water 
level of the supply. Thus, referring to Fig. 19, the bent pipe 
ABCD connects vessels M and N, and the highest point B of the 
pipe is higher than A, the level of the liquid in M. Assuming 
that the liquid is water, water will flow from M through the pipe 
to N as long as the water level in M is higher than in N. A 
siphon will not start itself, but when the part ABC of the siphon 
pipe is filled with water, water will flow from M to N until the 



§4 



PNEUMATICS 



51 



vessel M is emptied or the water level is the same in both vessels. 
The reason that the water rises in the part AB of the siphon is 
that when the water falls in the part BD, it leaves a partial 
vacuum behind it, and the atmospheric pressure on the water at 
A forces the water up the part AB. Evidently, the height EB 
= A' must not exceed 34 ft. ; in practice, it is not advisable to 
have it exceed 28 ft. The height FE = his the difference of level 
of the liquid in the two vessels. If h is in feet, A X H is the 
specific pressure in pounds per square inch that urges the water 
from M to N; because BF X ^^ = specific pressure at D 
due to the water in BD, and this is decreased by the partial 
vacuum represented by EB X ^f , thus leaving an active 



specific pressure of {BF — BE) \-^ = 



hX 



1 3 

Sir- 




If the pipe is a small rubber tube, the siphon may be started 
when the height h' is not too great by placing the end D in the 
mouth and sucking out the air. If the water level in both vessels 
is in the same horizontal plane, the siphon will not v/ork, since 
there will then be no specific pressure due to the difference in 
the water levels. The action of a siphon may be interrupted 
and resumed by means of a valve or clamp on the delivery pipe. 

85. Barometers. — The word barometer is derived from the 
Greek, and literally means weight measurer; it is an instrument 
used for measuring the pressure of the atmosphere. There are 
two forms in common use: the mercurial barometer and the 
aneroid barometer. 



52 



ELEMENTS OF PHYSICS 



§4 



A common form of mercurial barometer is shown in Fig, 20. 
It consists of a glass tube A, closed at the upper end, with the 
lower end, which is open, inserted in a cup of mer- 
cury C, the arrangement being similar to that shown 
in Fig. 17. The tube and cup are attached to a 
wooden frame F, to which is also attached a scale S 
and an accurate thermometer T. The scale S car- 
ries a vernier, by means of which (in the instrument 
shown) readings may be taken to }^{o mm., or about 
.004 in. Whenever a reading of the barometer is 
taken, a reading of the attached thermometer is 
also taken, and then, by means of tables or by cal- 
culation, if accuracy is desired, a correction is made 
of the reading to reduce it to 0° C. (= 32° F.). 
Another correction is made to reduce the reading to 
sea level, and a third correction is made for capil- 
larity. When all these corrections have been made, 
the final result is called the barometric pressure, 
and it represents the pressure of the atmosphere, 
under standard conditions, at the time the reading 
was taken, the pressure being in millimeters or 
inches of mercury, according to how the scale is 
graduated. 



Fig. 




Fig. 21. 



86. Aneroid barometers are made in many forms, some being of 
the shape and size of a watch ; they contain no liquid. A common 



§4 PNEUMATICS 53 

form is shown in Fig. 21 ; it is a metal box, circular in shape, air- 
tight, and carrying a glass face and dial, the latter being graduated 
in inches and tenths. The air is exhausted from the box, and the 
back is so constructed and fastened to a set of multiplying levers 
that a very slight movement of the back will result in a large 
movement of the hand that passes over the face of the dial. An 
increase in the air pressure presses the back in slightly and moves 
the hand ; a decrease in the air pressure causes the hand to move 
the other way. These barometers are compensated for tempera- 
ture; and they are frequently so made that they can be adjusted , 
for any particular altitude so as to read for sea level. When 
well made, they are very accurate, and some will show a differ- 
ence in pressure for a vertical height of only a few feet, say be- 
tween the floor and the top of a table. 

87. Barometers are employed for three purposes: to show the 
altitude of a place above sea level or the difference of altitude 
between two places; to show the barometric pressure at any 
place at some particular time; and to indicate changes in the 
weather. In connection with the latter purpose, it is to be re- 
membered that air containing water vapor is less dense than dry 
air; hence, if several observations at regular intervals throughout 
the day show a falling barometer, the moisture content (humid- 
ity) of the air is increasing, and if this continues, a storm is 
probable. A rising barometer, on the contrary, shows that the 
air is becoming dryer, and fair weather is probable. 

88. Absolute Pressure. — Suppose a large book, say a diction- 
ary, weighing 20 pounds to rest on top of a table, the pressure on 
the table due to the dictionary is 20 pounds. But there is also a 
pressure on top of the table (and the book) due to the atmosphere. 
The sum of the two pressures is the total pressure on the top of 
the table, and is called the absolute pressure. The absolute 
pressure, then, is the sum of the total external pressures, which 
may be called the apparent pressure, on any surface and the 
atmospheric pressure on that surface. Ordinarily, only the 
apparent pressures need be considered, since the atmospheric 
pressure in any direction is balanced by an equal (or practically 
equal) pressure in the opposite direction. Thus, in the case of the 
table, the downward pressure on the upper surface of the top is 
balanced by an upward pressure on the lower surface, and the 
pressures on the edges are also balanced. Therefore, insofar as 



54 ELEMENTS OF PHYSICS §4 

any movement of the table is concerned, the atmospheric pressure 
has no effect. 

Absolute pressures begin at 0, called zero absolute, which is 
the pressure in a perfect vacuum, and they increase upward in- 
definitely. There can be no such thing as a negative absolute 
pressure. 

89. Pressure Gauges. — Pressures of liquids and gases are 
usually measured by means of instruments called gauges, and 
pressures recorded on such instruments are called gauge pressures. 
The dial of a gauge is always so graduated that the zero point 
corresponds to the pressure of the atmosphere only on the fluid; 
in other words, to the pressure on the fluid when acted on only 
by the atmosphere. Consequently, the absolute pressure is 
always equal to the gauge pressure plus the barometric pressure. 
If the barometric pressure is not known, add 14.7 to the gauge 
pressure, if the pressure is recorded on the dial in pounds per 
square inch. In what are called vacuum gauges, which record 
pressures below 0, gauge pressure, the scale readings are in 
inches of mercury, and when converted into pounds per square 
inch, pressures in the partial vacuum (see Art. 79) , are absolute 
pressures. Gauges are also made to read in metric units, as 
milhmeters of mercury or pressure per square centimeter. 

Pressures on solids and liquids are nearly always taken as 
pressures above the atmosphere, i. e., as gauge pressures; in the 
case of gases, however, as will presently be shown, it is frequently 
necessary to use absolute pressures. 



EXAMPLES 



(1) Convert (a) 100°F. into degrees centigrade; (b) — 40°C. into degrees 
F.; (c) 1265°C. into degrees Fahrenheit. [ (a) 37%°C. 

Ans. I (&) -40°F. 
[ (c) 2309°F. 

(2) What is (a) the tension of the steam in a condenser when the vacuum 
is 24.4 in.? (6) A tension of 12.5 lb. per sq. in. is equivalent to how many- 
inches of vacuum? 4 / ^^^ 2.712 lb. per sq. in. 

\ (6) 4.47 in. of vacuum. 

(3) The diameter of a pump piston is 6 in., its stroke is 9 in., and the 
total height of lift is 52 ft. 9 in.; what is the work done during one stroke? 

Ans. 521.5 ft.-lb. 



ELEMENTS OF PHYSICS 

(PART 1) 



EXAMINATION QUESTIONS 

(1) If a body be weighed on a spring scale in Boston and also 
in Baltimore, in which place will it weigh the more and why? 

(2) A certain body weighs 1 ton = 2000 pounds on a beam 
scale that has been standardized to g = 980.665; how much will 
it weigh on a spring scale in San Diego that has been standardized 
to the value of g at that place, if </ = 32.1373? 

Ans. 1997.71 lb. 

(3) A body weighs 500 pounds and has a velocity of 40 ft. per 
sec. Due to the action of a steady force, the velocity is increased 
to 75 ft. per sec. in 3 seconds. What is (a) the acceleration? 
(6) what force is required to produce this change in velocity? 

^«o / <^^) 1^3 ft.persec.2 
^^^- I (6) 181.4 lb. 

(4) A stone block (ashlar) is 8 in. high, 14 in. wide, and 9 ft. 
long. If its specific gravity is 2.5 what is (a) the weight of the 
block? what is the density of the stone? . f (a) 1092 lb. 

^''^' I (6) 4.85 

(5) The plunger of a hydraulic jack on which the load is sup- 
ported is 4 in. in diameter; the piston that forces the water or 
other liquid into the jack has a diameter of | in.; what load will 
the jack lift, if the pressure on the piston is 55 lbs? 

Ans. 3520 lb. 

(6) Referring to the last example, if the piston moves 3i in. 
during each stroke (a) how many strokes are required to raise 
the load 2f in.? (6) how much work would be done in doing this? 

(a) 50 strokes. 



^'^^- ' (6) 80.67 foot-lb. 
(7) A specimen of a certain alloy weighs 2 lb. 3 oz. in air and 
1 lb. 13f oz. in water; what is the specific gravity of the alloy? 

Ans. 8.667. 
&A 55 



56 ELEMENTS OF PHYSICS §4 

(8) Referring to the last example, what was the volume of the 
specimen in cubic inches? Ans. 9.09 cu. in. 

(9) How many degrees Fahrenheit are equivalent (a) to 
—120° C? to 625°C.? . j (a) — 184°F. 

1(6) 1157°F. 

(10) Comparatively low pressures are sometimes measured 
by means of an instrument called a manometer, which records 
the pressure in inches of mercury. If the manometer reads 
45.83 in. of mercury, what is (a) the equivalent pressure in 
pounds per square inch? (6) in atmospheres? 

, f (a) 22.51 lb. per sq. in. 

1 (b) 1.531 atmospheres. 

(11) Convert (a) 1859°F. and — 210°F. into degrees centigrade. 

(a) 1015°C. 



^^«- ' (6) _135°C. 

(12) What is the pressure in a condenser when the vacuum 
gauge reads 22 in.? Ans. 3.89 lb. per sq. in. 

(13) The plunger of a steam pump is 2| in. in diameter and its 
stroke is 5 in.; the height of suction is 17 ft.; the level of discharge 
is 26 ft. above the pump; and the pump makes 240 working 
strokes per minute. How much work must be done per minute 
by the pump? Ans. 9147 ft.-lb. 

(14) The specific gravity of a certain liquid is 1.52 and of 
another liquid .977; (a) what would be the hydrometer reading 
of the first liquid in degrees Beaume? (6) of the second liquid? 

^'''- 1(6) 13.3°Be. 

(15) The hydrometer reading for a certain liquid that is 
lighter than water is 26° Be., and for another liquid that is 
heavier than water the reading is also 26° Be.; what is (a) the 
specific gravity of the first liquid? (6) of the second liquid? 

^''^' 1(6) 1.2185. 



ELEMENTS OF PHYSICS 

(PART 2) 



PNEUMATICS 

(Continued) 



PROPERTIES OF PERFECT GASES 

90. Boyle's Law. — While a theoretically perfect gas does not 
exist, air, hydrogen, nitrogen, oxygen, and many other gases 
may be considered as perfect gases in ordinary practical appHca- 
tions, and Boyle's law (also called Mariotte's law) may be used 
in all such cases. This law may be stated as follows: 

The temperature remaining the same, the volume of any perfect 
gas varies inversely as its absolute pressure. 

Let pi and vi be the absolute pressure and the volume of some 
perfect gas, say air, at a temperature of t°; suppose the pressure 
is changed to p2, then, the temperature being still t°, the volume 
V2 may be found from the proportion, 

Pi '■ P2 = V2 : Vi 
in accordance with Boyle's law. From this proportion, 

PlVi = P2V2; 
and, in general, 

p^vi = P2V2 = P3V3 = etc. = k 
in which A; is a constant. 

Example. — If the volume of air in an air compressor is 1.82 cu. ft. at atmos- 
pheric pressure (14.7 lb. per sq. in.) and it is compressed to 38 lb. per sq. 
in., gauge, what is its volume, the temperature being the same in both 
cases? 

Solution. — The absolute pressure before compression is 14.7 and after 
compression, 38 + 14.7 = 52.7 lb. per sq. in. Substituting in the above 

formula, 14.7 X 1.82 = 52.7 X v^, or % = ^'^''^J^J-^^ = .5077 - cu. ft. 
Ans. 

91. — In accordance with Boyle's law, if the pressure be doubled, 
the volume will be halved, or if the pressure be halved (by ex- 

§4 57 



58 ELEMENTS OF PHYSICS §4 

pansion of the gas) , the volume will be doubled. In gieneral, if 
the pressure be increased (or decreased) n times, the volume will 

be decreased (or increased) — times. This will be evident from 

the formula; because, if p2 = npi, piVi = npiV2 and ^2 = -Vi. 

Example. — Suppose air at a gauge pressure of 45 pounds per sq. in. and 
occupying a space of 984 cu. in. is allowed to expand to 4 times its volume. 
If the barometric pressure is 30.106 in., what is the pressure, gauge, after 
expansion, the temperature remaining the same? 

Solution. — The absolute pressure before expansion is 45 + 14.7 = 59.7 
lb. per sq. in. Since the volume increased 4 times, the pressure after ex- 
pansion is 59.7 -j- 4 = 14.925 lb. per sq. in., absolute, and the gauge pres- 
sure is 14.925 — 14.7 = .225 lb. per sq. in. Ans. 

The same result might have been obtained by applying the formula of 
the last article. Thus, the volume after expansion was 984 X 4 = 3936 cu. 

in.; hence, 59.7 X 984 = p2 X 3936, and p2 = —^ofg — = 14.925. 

92. The density and specific weight (weight of a unit of volume) 
vary directly as the pressure and inversely as the volume, the 
temperature remaining the same. This is a direct consequence of 
Boyle's law, as a little consideration will show. Suppose the 
volume of .256 pound of air at a temperature of t° and a pressure 
of 14.7 lb. per sq. in., absolute, is 3.2 cu. ft. The density and 
specific weight of a gas are taken as equal numerically; and the 

.256 
density in this case is -w-^ = .08. Now, if this air be allowed to 

expand to, say, 3 times its original volume, it will occupy a space 

256 
of 3.2 X 3 = 9.6 cu. ft., and its density will be ^ = .02%, 

(which is 3^^ the original density. Or, if the pressure be increased 

n times, the volume will be - times as large, and the density will 

be n times as great. Let Wi and Wi be the specific weights 
(densities) of the gas, vi and V2 the corresponding volumes, and 
pi and 2?2 the corresponding pressures, absolute; then, 

WiVi = W2V2 (1) 

and WxPi = W2P1 (2) 

Example. — The weight of 1 liter of air at 0° C. and a pressure of 1 atmos- 
phere is 1.2929 grams; what is the weight of 15 cu. m. when the pressure is 
4.6 atmospheres, both pressures being absolute? 

Solution. — Applying formula (2), Wi = 1.2929, pi = 1, p^ = 4.6, and 
1.2929 X 4.6 = u;2 X 1, or w^ = 1.2929 X 4.6 = 5.94734 grams per liter. 



§4 



PNEUMATICS 



59 



Since 1 liter = volume of 1 cu. dm., and 1 cu. m. = 1000 cu. dm., 15 cu. m. 
at 0° C. and under a pressure of 4.6 atmospheres weigh 

15 X 1000 X 5.94734 = 89210 g. = 89.21 Kg. Ans. 

93. Gay-Lussac's Law. — Referring to Fig. 22, C is a cylinder, 
closed at one end and open at the other, and containing a tightly 
fitting piston P on which is laid a weight W. The space beneath 
the piston is filled with air under atmospheric pressure. When 
the piston is released, the weight of the piston and of the load W 
increases the specific pressure on the confined air; the piston 
moves downward, compressing the air until its tension exactly 
equals the specific pressure of the atmosphere plus the specific 
pressure due to the piston and 
weight W. Suppose the area of 
the piston is 25 sq. in., its weight 
is 10 pounds, and weight of W 
is 20 pounds. The specific pres- 
sure due to weight of P and W is 

10 + 20 

per sq. 



yy/////^/ 



25 



1.2 lb. 



m. 



W 



w 



Ml 1'^ 



■illlllNIIIIIIIIIIlT 



h' 



'A ,, 

iiiiiiiiiiiiil 



Fig. 22. 



and the tension of the confined 
air, assuming the barometric 
pressure to be 14.7 pounds per 
sq. in., is 14.7 + 1.2 = 15.9 
lb, per sq. in., absolute. If, 
therefore, the volume under- 
neath the piston before it was released be represented by Vi and 
after release by ^2, the change in volume may be represented by 
the distance that the piston moves down. For, if A is the 
distance between the bottom of the cylinder and the bottom 
of the piston, before release (indicated by the dotted lines) and 
K is the position after release, ui = 25 X A and ^2 = 25 X h! ^ and 
Vx 2bh h ^. . . . , 

— = 2Kp = X7- Iheretore, any change m the tension of the 

confined air will produce a change in volume that may be meas- 
ured by — will be proportional to — the distance moved through 
by the piston up or down. It is assumed, for convenience, that 
the piston moves without friction and that any change in the 
tension of the air, however slight it may be, will cause a move- 
ment of the piston. 

Suppose the temperature of the confined air is 0° C. (32° F.), 
as shown by a thermometer placed within the cylinder and that 



60 ELEMENTS OF PHYSICS §4 

heat is applied to the cyHnder, It will be found that for every 
degree rise in temperature centigrade, indicated by the ther- 
mometer, the piston rises ^T^^rd of the distance /i' ; that is, for every 
degree rise in temperature, the volume increases TTsrd part. 
Note that the tension of the air is not changed, since the specific 
pressure is the same as before. The heat has increased the 
vibratory motion of the molecules of air and caused the air to 
expand. If the cylinder be cooled, so that the temperature of the 
air is less than 0° C, the piston falls xr^rd of the distance h' for 
every degree fall in temperature. Therefore, 

When the pressure remains the same, the volume varies directly as 
the temperature. 

Now suppose the piston be fixed so it cannot move ; the volume 
then remains constant. Attach a gauge to the cylinder to meas- 
ure the tension of the confined air. On heating the cylinder, it 
will be found that for every degree rise in temperature, the pres- 
sure (tension) of the air, as shown by the gauge, increases ^y^d; 
and for every decrease of 1° C. in temperature, the pressure falls 
^^d. Therefore, 

When the volume remains the same, the pressure varies directly as 
the temperature. 

These two statements may be expressed in the following 
general law: 

The pressure and volume of a confined gas vary directly as the 
temperature, when either volume or pressure, respectively, is constant. 

This is commonly called Gay-Lussac's law; it is also frequently 
called Charles' law, being named after two French scientists who 
discovered it. 

94. Since 1° C. = T = 1.8° F., and 273 X 1.8 = 491.4, it 
follows that for every degree rise in temperature indicated by a 
Fahrenheit thermometer, the volume or pressure will increase 

jgT^ = .002035th, and vice versa, according as the pressure or 

volume is constant. 

Let V = the volume of the air, p = tension of the air, t 
= number of degrees change in temperature, and c = the in- 
crease in volume or pressure due to 1° rise in temperature; then, 
when the pressure is constant, 

Vi = V -\- cvt = v(l + ct) ; (1) 

and when the volume is constant, 

Pi = P + cpt = p(l + ct). (2) 



§4 PNEUMATICS 61 

If a centigrade thermometer is used, c = yfj = .003663; if a 
Fahrenheit thermometer is used, c = .002035, Note that if the 
temperature be increased 273° C. = 491.4° F., the volume or 
pressure will then be doubled, since the quantity in parenthesis, 
1 + ct, then becomes 1 + ^ X 273 = 1 + 1 = 2. 
If the temperature falls, formulas (1) and (2) become 
v\ = V — cvt = v{l — ct), (3) 

and 

Pi = p — cpt = p[l — ct). (4) 

Example 1. — If the temperature of 3.42 cu. m. of air is raised 86° C. under 
constant pressure, what is the volume? 

Solution. — Applying formula (1), v = 3.42 and t = 86; whence, Vi 
= 3.42(1 + .003663 X 86) = 3.42 X 1.31502 = 4.4974 cu. m. Ans. 

Example 2. — A certain amount of air is cooled at constant volume from 
32° F. and a tension of 45.46 lb. per sq. in. absolute, to —92° F.; what is the 
tension after cooling? 

Solution. — Applying formula (4), p = 45.46 and t = 32° — (—92°) 
= 124°; whence, pi = 45.46(1 - .002035 X 124) = 45.46 X .74766 = 33.989 
lb. per sq. in. Ans. 

95. Absolute Temperature. — Referring to formulas (3) and 
(4) of the last article, t =^ to — ti, in which to is always the tem- 
perature at 0° C. or 32° F. Strictly speaking, t in formulas (1) 
and (2) is always equal to ^i — to. With this understood, it will 
be seen that if t = 273° C. in formula (3), wi = z;(l — ^tt X 
273) = 0, that is, the volume disappears; this, of course is im- 
possible, since air is matter, and matter cannot be destroyed. 
If, however, 273 be substituted for t in formula (4), pi = 0, that 
is, the air has no tension — its molecules have ceased to vibrate. 
For this reason, a temperature of 273° C. below 0° C. is called 
absolute zero of temperature. Since temperature is a measure 
of the vibrations of the molecules of a body, just as weight is a 
measure of the earth's attraction on a body, it is assumed that 
at absolute zero = —273° C, there is no molecular movement. 

The temperature indicated by the thermometer is usually 
indicated by t, and the temperature above 0°, absolute, called the 
absolute temperature, is indicated by T. The absolute tempera- 
ture centigrade is therefore equal to T = t -\- 273, usually 
written 273 + t. Thus, the absolute temperature of the boiling 
point of water at sea level, with the barometer at 760 mm., is 
273 + 100 = 373° C. 

Since 273° C. = 491.4° F., and 0° C. = 32° F. , absolute zero 
Fahrenheit = -491.4° + 32° = -459.4°; that is, absolute zero 



62 ELEMENTS OF PHYSICS §4 

on the Fahrenheit scale is 459.4° below of the Fahrenheit ther- 
mometer. To find the absolute temperature on the Fahrenheit 
scale, add 459.4 to the reading of the Fahrenheit thermometer; 
thus, the boiling point of water has an absolute temperature of 
459.4 + 212 = 671.4° F. = 671.4 X | = 373° C. 

If the temperature is below zero on either scale, the absolute 
temperature will be found in exactly the same manner. Thus, 
25° below zero, centigrade = - 25° C, and 273° + ( - 25°) 
= 273° - 25° = 248° C, absolute; and -25° F. = 459.4° - 25° 
= 434.4° F., absolute. 

96. The quantity in parenthesis in formulas (1) — (4), Art. 

94, may be written 1 + ^73^ = 1 + 273 = — 273 ^ 973' 

f 273 t T 

^ ~ 273 ^ — 273 — "^ 273" Consequently, formulas (1) and 

(3) may be written 

and formulas (2) and (4) may be written 

Vi-v{^ -cpT, (2) 

the value of c being the same as in the preceding formulas, and T 
being the absolute temperature after heating or cooling. These 
two formulas assume that the original temperature of the gas is 
273° C. = 491.4° F. If the original temperature is Tg and the 
final temperature Ti, then 

Vi:vo= Ti: To 

and pi :po = Ti : To 

from which 

VoTi , . 

Vl = -7p- (3) 

-t 

and 

Vl = ^ (4) 

Applying formula (2) to the second example of Art. 94, To 
= 459.4 + 32 = 491.4, Ti = 459.4 - 92 = 367.4, po = 45.46 and 

Pi = — AQ^ A — ~ ~ 33.989 lb. per sq. in., the same result as 

was previously obtained. 



§4 PNEUMATICS 63 

Example. — If a constant volume of gas is heated from 60° F. to 340° F., 
and the final tension is 36 lb. per sq. in., gauge, what was the original 
tension? 

Solution.— To = 459.4 + 60 = 519.4, Ti = 459.4 + 340 = 799.4, pi 

799 4 
= 36 + 14.7 = 50.7. Applying formula (4), 50.7 = ^V^ X po, or po 

50 7 X 519 4 
= ' ygg^ = 32.9 lb. per sq. in., absolute = 32.9 - 14.7 = 18.2 lb. 

per sq. in., gauge. Ans. 

It is useless to obtain more than 3 significant figures when the barometric 
pressure is assumed as 14.7 pounds per sq. in. 

97. Combined Law of Boyle and Gay-Lussac.^ — Boyle's law 
and Gay-Lussac's law may be combined into a single law, as 
follows: 

The product of the pressure and volume of any perfect gas varies 
directly as the absolute temperature. 

Let po, vo, and To be the original pressure, volume, and 
temperature, pi, vi, and Ti the pressure, volume, and temperature 
after a change; then 

2^° = 2^ = 2|^^ = etc. = iJ (1) 

in which R is a, constant for any particular gas, but has different 
values for different gases. To find the value of R for air, write 

PoVo „ 

-I 

Here po is the absolute specific pressure = 14.6946 lb. per 
sq. in. at sea level, with the barometer at 29.9213 in., and the 
thermometer at 32° F., for which To = 32° + 459.4° = 491.4°. 
To fix the value of Vo, note that the volume of 1 pound of air 

(the specific volume) under the above conditions is ^p^-., ; 

since 1 cu. ft. of air under these conditions weighs .08071 lb. 

Substituting these values in the last equation, 

14 6Q46 V 1 

49L4 X .08071 ^ ^' '''" ^ ^ -370506, say .37051 

pv 
Consequently, for air, ~ = .37051, or 

pv = .37051 T (2) 

Formula (1) can be applied only when the weight of the air is 1 
pound; to make it applicable to any weight of air w \h., mul- 
tiply the right-hand member of the equation by w, and 

pv = .37051 wT (3) 



64 ELEMENTS OF PHYSICS §4 

To find the value of i^for any gas other than air, divide, . 370506 
by the specific gravity of the gas. Thus, the specific gravity of 
hydrogen is .06926, and the value of R for hydrogen is 
.370506 -r- .0,6926 = 5.3495; the specific gravity of oxygen is 
1.10535, and the value of R for oxygen is .370506 -^ 1.10535 
= .33519. 

In formulas (2) and (3) , p is the pressure (absolute) in pounds 
per square inch, v is the volume in cubic feet, T is the absolute 
temperature, Fahrenheit, and w is the weight of the air in pounds. 
In formula (1), vo, vi, etc., may be expressed in any units — cubic 
feet, cubic meters, etc. 

Example 1. — What is the weight of 18,000 cubic feet of air at a tempera- 
ture of 60° F. and a tension of 26.3 lb. per sq. in., gauge? 

Solution. — The absolute pressure is 14.7 + 26.3 = 44 lb. per sq. in. 
and the absolute temperature is 459.4 + 60 = 519.4°. Substituting in 
formula (3), 44 X 18000 = .37051 XwX 519.4; whence, 

^ = :3^^5^ = ^^^^-^^^- "^^^- 

Example 2. — The volume of the cylinder of an air compressor is 28,276 
cu. in. If the reading of the barometer, corrected, is 30.1 inches, and the 
temperature of the air is 56° F., it can be shown that if the air be compressed 
to one-fourth its volume without losing any heat by radiation or otherwise, 
the pressure (tension) will be 104.4 lb. per sq. in., absolute; what will be 
its temperature? 

Solution. — Since the volume, pressure, and temperature all change, 
apply formula (1). Here po = the pressure of the atmospheric air = 30.1 
X .49111 = 14.782 lb. per sq. in., Vo = 28,276 cu. in., To = 459.4 + 56 
= 515.4, pi = 104.4 lb. per sq. in., and vi = 28,276 XH = 7069 cu. in. 

^, , , , . ,,, 14.782 X 28276 104.4 X 7069 . 
Therefore, by formula (Ij, -^ _ . = ^ , from which, 

104 4 V 7069 V 51 ^ 4 
Tx = 14 782 X 2^76 ^ ^^^° absolute = 910 - 459.4 = 450.6° F. 

Ans. 

Example 3. — What is the specific volume of atmospheric air when the 
barometer stands at 29.652 in. and the temperature is 100° F.? 

Solution. — The specific volume is the volume of 1 pound in cubic feet; 
hence, applying formula (2), p = 29.652 X .49111 = 14.562 lb. per sq. 
in., T = 459.4 + 100 = 559.4, and 14.562 Xv ^ .37051 X 559.4; from 

... .37051 X 559.4 ^^ _„ .^ . 

which, V = ^ . cgr. = 14.233 cu. ft. Ans. 

' 14.562 

98. Mixture of Gases. — Suppose two liquids of different den- 
sities to be mixed and that the mixture is allowed to stand. If 
there be no chemical action between the liquids, it will be found 
that after a time the liquids have separated, the lighter floating 
on the heavier. The case is entirely different with gases, since 



§4 PNEUMATICS 65 

if two vessels containing gases of different densities are connected 
with each other, it will be found after a time that the two gases have 
mixed, the density of the mixture in both vessels being the same; 
even though the vessel containing the lighter gas be placed over 
the other, the same result will be obtained — the density and 
quality of the mixture of the gases in both vessels will be the same. 
This is an extremely useful property of gases ; otherwise no life 
of any kind could exist. As an example of this fact, it must be 
borne in mind that when anything is burned or when any animal 
or vegetable matter decays, carbon dioxide gas is given off, it be- 
ing a product of combustion and decay, which is slow combustion. 
Since carbon dioxide is more than 13^^ times as heavy as air, it 
follows that if gases separated in accordance with their densities, 
as in the case of liquids, the carbon dioxide would form a layer 
covering the entire earth, and would kill all forms of animal life. 

Rule. — // two or more gases are mixed, the product of the pressure 
and volume of the mixture is equal to the sum of the products of the 
pressures and volumes of the gases, provided there is no chemical 
action and the temperature of the gases is the same. 

Let P and V be the pressure and volume of the mixture, pi 
P2, Ps, etc., the pressures of the gases, and vi, v^, Vz, etc., the corres- 
ponding volumes; then 

PY = pxvx + PiVi + pzv% + etc. 

Example. — Suppose two vessels to contain air at the same temperature. 
The volume of one vessel is 5.72 cu. ft. and the tension of the air is 17 lb. 
per sq. in., gauge; the volume of the other vessel is 8.36 cu. ft. and the ten- 
sion of the air is 24 lb. per sq. in., gauge; if they are allowed to communi- 
cate with a third vessel (empty) having a volume of 3.48 cu. ft., what will be 
the pressure of the mixture? 

Solution. — The volume of the mixture is 5.72 + 8.36 + 3.48 = 17.56 
cu. ft., since if both vessels communicate with the empty vesssel, the air 
can mix in all three vessels. Then, applying the formula, the absolute 
pressures are 14.7 + 17 = 31.7 and 14.7 + 24 = 38.7, and P X 17.56 

= 31.7X5.72 +38.7 X 8.36 = 504.856, from which, P = ^^^ff^ = 28.75 

17.66 

lb. per sq. in., absolute = 28.75 — 14.7 = 14.05 lb. per sq. in., gauge. Ans. 
If the gases are alike and the temperatures are different, 
then, letting T, T\, T2, etc., be the corresponding absolute tem- 
peratures, 

PV _ PlVl , p^ , psvs , , 

rp m I rp ~T~ rp t etC. 

Example. — In the example of the last article, suppose the temperature 
of the air in the smaller vessel had been 95° F., in the larger vessel 50° F., 
5 



66 ELEMENTS OF PHYSICS §4 

and the temperature of the mixture after a time was found to be 62° F. ; 
what was the tension of the mixture at that time. 

Solution. — The absolute temperatures are T = 459.4 + 62 = 521.4°, 
Ti = 459.4 + 95 = 554.4, and T2 = 459.4 + 50 = 509.4. Applying the 
PX 17.56 31.7 X 5.72 38.7 X 8.36 

formula, g21.4 " 554:4 ^ 6091 " -96219, from which 

KOI 4 V Q621Q 
p = 17 56 = 2^-^^ ^^- P^^ ^^- ^^•' absolute = 28.57 - 14.7 

= 13.87 lb. per sq. in., gauge. Ans. 



EXAMPLES 

(1) Referring to Fig. 22, if 1.86 cu. ft. of air under a gauge pressure of 
51.51b. per sq. in. be allowed to expand to 4.28 cu. ft. by decreasing the 
weight W, what will be the pressure if there is no change in the tempera- 
ture? Ans. 13.94 lb. per sq. in., gauge. 

(2) Referring to Art. 97, what is the weight of 220,000 cu. ft. of hydrogen 
at 75°F. under a pressure of 1 atmosphere? Ans. 1130.8 lb. 

Note. — "When stated in this manner, the pressure is always understood to mean absolute 
pressure and under standard conditions i.e., 14.6946 lb. per sq. in. 

(3) A certain volume of air has a tension of 21.7 lb. per sq. in., gauge, 
and a temperature of 265°F.; it is cooled until the temperature is 62°F. ; 
what is the tension? Ans. 11.93 lb. per sq. in., gauge. 

(4) A vessel containing 750 cu. in. of free air has forced into it 2120 cu. in. 
of air under a pressure of 52.63 lb. per sq. in., gauge. If the tempera- 
ture of both volumes of air and of the mixture is the same and the baro- 
metric pressure is 30.6 in. of mercury, what is the tension of the mixture? 

Ans. 191.25 lb. per sq. in., gauge. 

Note. — "When the barometricpressure is given, it must be used instead of 14.7 or 14.6946 
lb. per sq. in. in finding the absolute pressure, and vice versa. 

(5) If, in the last example, the temperature of the free air had been 50° 
F., and that of the air in the other vessel 208°, what is the tension of the 
mixture when its temperature is 92°? Ans. 158.51 lb. per sq. in., gauge. 



HEAT 



NATURE AND MEASUREMENT OF HEAT 

99. Nature of Heat.— Heat may be defined as molecular energy. 
The molecules of a body have mass, and since they are in rapid 
motion, they possess kinetic energy, "which manifests itself in the 
phenomenon called heat. At the temperature of absolute zero, 
all movement of the molecules ceases, the body posessess no 
heat, and is absolutely inert — dead. For any temperature above 
0°, absolute, the body has an amount of heat that is proportional 
to the temperature, the temperature being dependent upon 



§4 HEAT 67 

the velocity of the molecules and being a measure of the amount of 
heat contained in the body. 

Heat is energy, and since energy manifests itself in different 
forms, heat may be changed into different forms of energy, such 
as mechanical energy, chemical energy, and electrical energy; 
conversely, these three forms of energy may all be converted into 
heat. 

100. Quantity of Heat. — The temperature of a body does not 
depend upon its mass or size. This is evident from the fact that 
if an ordinary pin and a ten-pound iron weight be placed in a tub 
of hot water, the pin, weight, and water will all have the same 
temperature after a time. In other words, temperature may be 
regarded as a measure of the specific molecular energy of a body; 
it is therefore independent of the size and shape of the body. 
At the same time, it is obvious that the iron weight contains a 
greater amount of heat (kinetic molecular energy) than the pin, 
and it is desirable to have some means of measuring heat in 
quantity, and this is accomplished by (a) measuring the amount 
of energy expended in raising the temperature, that is, by chang- 
ing another form of energy into heat; (6) measuring the energy 
produced by a fall of temperature, when changing heat into some 
other form of energy. 

The most obvious way of measuring heat is by measuring the 
work required to overcome friction. Assuming that there is no 
wear or that the wear is so exceedingly small that it can be dis- 
regarded, all the work of friction is converted into heat. Taking 
as a standard for one heat unit the work of friction required to 
heat one pound of water from 63°F. to 64°F., it has been found by 
careful and elaborate tests that the amount of work required for 
this purpose is very closely 778 foot-pounds, which is called the 
mechanical equivalent of heat. When the centigrade scale is 
used, the work required to raise 1 kilogram of water from 17° 
to 18°C. (62.6°F. to 64.4°F.) is very nearly 427 meter-kilograms. 

101. The British thermal unit (abbreviation, B.t.u.) is the 
amount of heat required to raise the temperature of one pound 
of water 1° F. (from 63° to 64°), and its mechanical equivalent 
is 778 foot-pounds. 

The calorie (abbreviation, cal.) is the amount of heat required 
to raise the temperature of one kilogram of water 1°C. (from 17° 
to 18°), and its mechanical equivalent is 427 m.-kg. 



68 ELEMENTS OF PHYSICS §4 

The reason for specifying the temperature is that the amount 
of heat required to raise the temperature of a given weight of 
water one degree is different for different temperatures. 

Since 1 kg. = 2.2046221b. and 1°C. = | = 1.8°F., 1 calorie is 
equal to 2.204622 X 1.8 = 3.96832 B.t.u. Also, 1 m.-kg. 
= 3.28084 X 2.204622 = 7.2330 ft.-lb., and 1 calorie = 778 

X ^^llp = 426.8 +, say 427 m.-kg. 

For many purposes, the calorie is too large a unit for convenient 
use, and what is called the small calorie or gram calorie is then 
employed; this unit is, the amount of heat required to raise the 
temperature of 1 gram of water from 17°C. to 18°C., and since 
a gram is 1-lOOOth of a kilogram, the gram calorie is 1-lOOOth of a 
large calorie (kilogram calorie), or 1000 gram calories = 1 kilo- 
gram calorie. The gram calorie is largely used by chemists, 
physicists, and scientists generally, while the kilogram calorie is 
used in engineering calculations. Both units are called calories, 
and the reader must be on his guard to distinguish which is 
meant; but a little familiarity with their use will soon show which 
is intended, since one is 1000 times as large as the other. 

102. Heat Capacity. — The capacity of a body for heat depends 
upon the weight (strictly speaking, the mass) of the body and 
upon the material of which the body is composed; for equal 
weights, the capacity depends only on the material composing 
the body. It can be shown by experiment that the number of heat 
units required to raise the temperature of equal weights of iron, 
lead, copper, wood, etc. 15°, say, are quite different; and when 
these substances are cooled, the amount of heat given off (which 
is exactly the same as that imparted in raising the temperature 
the same amount the body is cooled) is also different. The more 
heat required to raise the temperature the greater is the capacity 
of the body for heat, and vice versa. 

103. Specific Heat. — The specific heat of a substance is the 
amount of heat required to raise the temperature of a unit 
weight of the substance 1° divided by the amount of heat re- 
quired to raise the temperature of the same weight of water 1° 
at some specified temperature. The specific heat of water is 
generally considered to be 1, in which case, the specific heat of 
any substance may be defined as the number of B.t.u. required 
to raise the temperature of the substance 1°. If the specific 
heat of a substance be denoted by c, the weight by w, and the 



§4 HEAT 69 

difference in temperature before and after heating (or cooling) 
by t, the number of heat units U (B.t.u.) required to heat the 
body t° (or given off in coohng ^°) is 

U = cwt 
For example, if the specific heat of lead is .0315, how many 
heat units are required to raise the temperature of 25 lbs. 
from 58°F. to 330°F.? here i = 330 - 58 = 272, w = 25, and c 
= .0315, therefore, the number of heat units required is U = .0315 
X 25 X 272 = 214.2 B.t.u., which is equivalent to 214.2 X 778 
= 166,647.6, say 166,600 foot-pounds of work. When multi- 
plying by 778 (or 427), the mechanical equivalent, it is useless to 
retain more than 3 or 4 significant figures in the product; this 
statement also applies to multiplying by the specific heat. 

104. The specific heat of most substances (if not all) is not 
constant, but varies somewhat with the temperature and with 
the pressure; for this reason, results obtained by application of the 
formula in the last article are not strictly accurate, but are 
sufficiently so for most practical purposes. The following table 
gives average values of the specific heats of some solids and liquids 
that may be used for practical purposes in connection with engi- 
neering calculations. Different authorities give slightly differ- 
ent values. Some of these values are useful in pulp mills, when 
in calculating the unit required to melt sulphur, etc. 

SPECIFIC HEATS 

SOLIDS 

Aluminum 218 Ice 504 Sulphur 180 

Brass 090 Lead 0315 Tin 055 

Cast Iron 130 Platinum 0325 Tungsten 034 

Charcoal 203 Steel (soft) 116 Wrought Iron 127 

Copper 095 Steel (hard) 117 Zinc 094 

Gold 032 Silver 056 Glass 199 

LIQUIDS 

Acetic Acid 51 Hydrochloric Lead (melted) 041 

Acid 60 

Alcohol (pure) 70 Mercury 033 Sulphur (melted) . . 235 

Benzine 43 Sulphuric Acid . . .336 Tin (melted) 058 

Glycerine 576 Water 1 Turpentine 472 

105. The specific heat of gases has two different values, ac- 
cording to whether the gas is heated at constant volume or under 
constant pressure. When heated at constant volume, all the 
heat goes to increasing the molecular energy, that is, the tempera- 



70 ELEMENTS OF PHYSICS §4 

ture; but when heated under constant pressure, the gas expands 
and does work, the work being equal to the increase in volume 
multiplied by the tension of the gas. Consequently, it requires 
more heat to raise the temperature of a gas under constant 
pressure than at constant volume. Letting Cp = specific heat under 
constant pressure and c» = specific heat at constant volume, the 
following table gives the specific heats of some gases. 

SPECIFIC HEATS 

GASES 

Cn Cy Cp Cy 

Acetylene 270 .024 Hydrogen 3.42 2.44 

Air 241 .171 Nitrogen 247 .176 

Carbon Monoxide. .243 . 172 Oxygen 217 .155 

Carbon Dioxide 210 .160 Sulphur Dioxide. .154 .123 

When the pressure and volume both change, as when a certain 
volume of compressed air expands and does work, the specific heat 
does not have either of the two values given above; the law that 
connects the pressure and volume under those circumstances is 
quite complicated, and will not be given here. 

106. Temperature of Mixtures of Different Bodies. — If two or 
more substances of the same material or of the same specific 
heats and having different temperatures are mixed, the tempera- 
ture of the mixture is readily found. Suppose, for example, 
that 10 pounds of water at 50° are mixed with 10 pounds of 
water at 86°; the temperature of one must be increased to the 
temperature of the mixture, and the temperature of the other 
must be decreased to the temperature of the mixture; then, since 
both bodies weigh the same, one receives as much heat as the 
other loses, and the temperature of the mixture will be midway 

50° 4- 86° 
between the temperatures of the two bodies, or ^ = 68°. 

Here one body has increased its temperature 68° — 50° = 18°, 
and the temperature of the other body has decreased 86° 
- 68° = 18°. 

Suppose, now, that 4 pounds of water at 50° are mixed with 
9 pounds of water at 86°; one body still receives as much heat as 
the other loses, but the temperature of the mixture will not be the 
mean of the temperatures of the two bodies, because the larger 
body has a greater capacity for heat than the smaller body; hence, 
any change in the temperature of the larger body will produce 
a greater change in the smaller body. Let x = the temperature 



§4 HEAT 71 

of the mixture; then, evidently, (86 —x)X 9 = {x — 50) X 4, 
since the larger body gives up (86 — re) X 9 heat units and the 
smaller body receives the same number, which must equal (x 
— 50) X 4 heat units, the specific heat of water being 1. The 
above equation reduces to 774 — 9x = 4:X — 200, which becomes, 
by transposition, 13a; = 974, or a; = 74.923°. This result 
might have been arrived at in the following manner: let tm and 
Wm be the temperature and weight of the mixture, ti and Wi the 
temperature and weight of one body, ^2 and W2 the temperature 
and weight of another body of the same material, etc.; then, 

Wmtm = Witi + W2t2 + Wzts + etC, (1) 

In the present case, «;„, = Wi + ^2 = 4 + 9 = 13, ^i = 50, 
and ti = 86; substituting in the formula, 

ISt^ = 4 X 50 + 9 X 86 = 200 + 774 = 974 
from which, t„, = 74.923° 

the same result as before. It will also be noted that the same 
operations are performed in both cases. 

If only the temperature of the mixture is desired, the above 
formula may be written. 

_ wi ^1 + W2^2 + Wsts + etc. ,„) 

"• ~ Wi-\-W2 + Ws + etc. ^ ^ 

107. If, however, the bodies mixed have different specific 
heats, every term in the righthand member of the above for- 
mula must be multiphed by the specific heat of the corresponding 
body, since the amount of heat given up or received by any body 
is cwt (Art. 103). and this is the same for all the bodies, including 
the mixture. In such case, 

_ CiWiti + C2W2t2 + CzWsts + etc. .„s 

"* "" CiWi + C2W2 + C3W3 + etc. 

Example 1. — Suppose a copper ball weighing 1.4 pounds and having a 
temperature of 860° is placed in a lead vessel weighing 3.8 pounds and con- 
taining 7.5 pounds of mercury, the temperature of the vessel and mercury 
being 74°; after all three bodies have arrived at the same temperature, with- 
out any loss of heat, what is the temperature of the mixture? 

Solution.— From the table of Art. 104. specific heat of copper is .095, 
of lead .0315, and of mercury .033. Substituting in the above formula, 

_ .095 X 1.4 X 860 + .0315 X 3.8 X 74 + .033 X 7.5 X 74^ 233° Ans 
^™ .095 X IT4 + .0315 X 3.8 + .033 X 7.5 

It is useless to use more than three (or at most, four) significant figures 
when the constants (specific heats) are given to only two or three significant 
figures. 

Example 2. — How many heat units (B.t.u.) are required to raise the 
temperature of a cast-iron ball weighing 28 lb. 4 oz. from 28°C. to 236°C.? 



72 ELEMENTS OF PHYSICS §4 

Solution.— Here f = 236 - 28 = 208°C. = 208 X 1.8 = 374.4°F. Us- 
ing the formula of Art. 103, c = .13, w = 28.25 pounds, and 

Z7 = .13 X 28.25 X 374.4 = 1375 B.t.u. Ans. 

Had the result been required in calories, change the weight to kilograms, 
obtaining 28.25 -^ 2.2046 = 12.814 kg. In this case, t = 208°C., and U 
= .13 X 12.814 X 208 = 346.5 cal. Or, since 1 cal. = 3.96832 B.t.u., 
1375 -^ 3.96832 = 346.5 cal. 



EXPANSION OF BODIES BY HEAT 

108. Kinds of Expansion. — 'When the temperature of the body- 
is changed its volume changes, an increase in temperature usually 
resulting in an increase in volume, an important exception being 
water, which decreases in volume from 32°F. to 39°F. and then 
increases for all higher temperatures. When the volume increases, 
the body is said to expand, and the change in volume is called 
cubic expansion. As the result of cubic expansion, the area of the 
surface and of any section is increased, and this change is called 
surface expansion; likewise, the length of any line is increased, 
and this change is called linear expansion. When a body is 
cooled, the opposite effect is produced, that is, the body contracts, 
which may be called negative expansion. Steel bridges generally 
have one end free, so they can move in the direction of the length 
of the bridge, since the bridge will be longer on a hot day in the 
summer than on a cold day in the winter; the difference in length 
may be an inch or more, depending on the length of the bridge 
and the difference in temperature. If both ends were rigidly 
fastened, this difference in length would tend to make the bridge 
buckle in the summer and tend to pull it apart in the winter. 

109. Coefficient of Expansion. — The increase in length for an 
increase in temperature of 1° between 32°F. and 33°F. divided 
by the original length at 32° is called the coefficient of linear 
expansion ; similarly, the increase in area divided by the original 
area, and the increase in volume divided by the original volume, 
under the same conditions, are called the coefficient of surface 
expansion and coefficient of cubic expansion, respectively. In 
the case of solids and liquids, the coefficient of expansion is so 
small that the coefficient of surface expansion is always taken as 
2 times the coefficient of linear expansion, and the coefficient of 
cubic expansion is always taken as 3 times the coefficient of linear 
expansion. 



§4 



HEAT 



73 



For every degree (Fahrenheit) rise in temperature, the body 
increases in length an amount equal to the product of the co- 
efficient (which is different for different materials) and the length 
at 32°, and when the temperature of a body decreases, the con- 
traction in length (negative expansion) is equal to the expansion 
of the body for the same rise in temperature. The coefficient of 
expansion varies somewhat with the temperature, but it is suffici- 
ently accurate in practice to multiply the length at the original 
temperature by the average value of the coefficient and by the 
difference in temperature to find the increase in length. Let 
ki, k^, and kz be the coefficients of linear, surface and cubical 
expansion, respectively; let ti be the original temperature and 
^2 the final temperature; let Zi and U, ai and ^2, and vi and V2 be 
the lengths, areas, and volumes at ti and ts, respectively; then, 

l2 = h + kill (h - tl) = Zl [1 + ^1 {t2 - ti)\ (1) 

^2 = ai [1 + ki (t2 - h)] (2) 

V2 = i^i [1 + A;3 (^2—^1)] (3) 

If the original temperature is higher than the final temperature, 
as when the body cools, the quantity in parenthesis, t2 — ti, will 
be negative and the final length, area, or volume will be less, as it 
should be. 

The following table gives for temperatures Fahrenheit the coef- 
ficients of expansion (average values) for a number of substances. 



COEFFICIENTS OF EXPANSION 



ki 



k. 



Aluminum 

Brass 

Copper 

Gold 

Iron (cast) 

Iron (wrought) . . 

Platinum 

Silver 

Steel 

Tin 

Zinc 

Mercury (60°F.). 
Alcohol (ethyl) . . 
Alcohol (methyl) . 
Gases (perfect) . . 



.00001234 

.00000957 

.00000887 

.00000786 

.00000556 

.00000648 

.00000479 

.00001079 

.00000636 

.00001163 

.00001407 

.00003333 

.000203 

.000267 

.000678 



.00002468 

.00001914 

.00001774 

.00001572 

.00001112 

.00001296 

.00000958 

.00002158 

.00001272 

.00002326 

.00002814 

.00006667 

.000407 

.000533 

.001357 



. 00003702 

.00002871 

.00002661 

. 00002352 

.00001668 

.00001944 

.00001437 

.00003237 

.00001908 

.00003489 

.00004221 

.00010000 

.000610 

. 000800 

.002035 



The coefficient of cubic expansion of gases is 1 -4- 491.4 = .002035. 



74 ELEMENTS OF PHYSICS §4 

The only liquids included are mercury and alcohol, because 
these two liquids are used in thermometers and because the 
coefficients of expansion for liquids vary considerably with the 
temperature. 

110. In machine shops, what are called shrinking fits are fre- 
quently used. For instance, the tires on locomotive drivers are 
"shrunk on." The part of the driver on which the tire is shrunk 
is called the wheel center; this is turned to a certain desired size, 
and the tire is bored a trifle smaller. The tire is then heated to 
quite a high temperature which causes it to expand sufficiently 
to slip over the wheel center. When cooled, the tire contracts 
and grips the wheel center with immense force. Crankpins, 
also, are frequently shrunk on in this manner. 

Example 1. — It is desired to- shrink a steel tire on a locomotive driver; 
if the diameter of the wheel center is 73.471 in. and the tire is bored to a 
diameter of 73.4 in., to what temperature (approximately) must the tire 
be heated to slip on the wheel center, assuming that the diameter when 
heated is .0013 in. larger than the wheel center? 

Solution. — Since the diameter is a line, the case is one of linear expansion, 
and formula (1) may be used. Here h = 73.4, h = 73.471 + -0013 
= 73.4723, fci = .00000636 (from table), and it is desired to find the 
difference of temperature k — h. Solving formula (1) for i2 — h, U — ti 

h - h 73.4723 - 73.4 ^ __„_, „ ^, . . , , ^ , ^, 

= —j-^ — = 70 4 y nonnOfi^fi ~ loo F. If the ongmal temperature of the 

tire is, say 70°F., the temperature after heating is 70 + 155 = 225°F. Ans. 

Example 2. — A round copper plate has a diameter of 8.336 in. at 62°F.; 
what will be the decrease in area when the temperature is 32°? 

Solution. — The original area is .7854 X 8.336^ = 54.5766 sq. in. Apply- 
ing formula (2), the area after cooling to 32° is (since k2 for copper 
= .00001774) 02 = 54.5766 [1 + .00001774(32 - 62)] = 54.5766 X .9994678 
= 54.5476 sq. in., and the decrease in area is 54.5766 — 54.5476 = .029 
sq. in. Ans. 

The same result might have been obtained somewhat more easily; thus, 
the decrease in area is evidently equal to hiait = ..00001774 X 54.5766 
X (62 - 32) = .029 sq. in. 

Example 3. — A silver vessel holds exactly 1 liter when the temperature 
is 20°C.; how many cubic centimeters will it hold when the temperature is 
100°C.? 

Solution. — Formula (3) must be used in this case, and ks for silver is 
.00003237; the difference in temperature is 80 - 20 = 60°C. = 60 
X 1.8 = 108°F. Therefore, since 1000 c.c. = 1 Uter = Vi, 

V2 = 1000(1 + .00003237 X 108) = 1003.5 c.c. = 1.0035 I. Ans. 

The result obtained in the last example shows how necessary 
it is to consider the temperature when accurate measurements 
of volume are desired. 



§4 HEAT 75 



HEAT TRANSMISSION 

111. Heat may be transmitted or propagated from one point or 
place to another point or place in three ways : by conduction; by 
convection; by radiation. 

112. Conduction. — If one end of an iron rod be heated, it will 
be found that after a time the other end is perceptibly warmer; 
and if the process continues, the end not in contact with the source 
of heat will become hot. The explanation is that the molecules 
at the heated end transmitted molecular energy (heat) to the 
adjacent molecules, these, in turn transmitted heat to the mole- 
cules adjacent to them, and so on, until the other end was reached. 
The longer the one end is in contact with the source of heat the 
hotter will the other end become; in other words, heat is con- 
ducted from one end to the other, passing from molecule to 
molecule. Again, if a vessel containing a fluid, say water, be 
placed in contact with a fire, as a kettle of water on a stove, the 
vessel becomes hot, and after a time, the water also becomes hot. 
In this case, two different substances are in contact — the metal of 
the vessel and the water in the vessel — and the heat is conducted 
through the metal, molecular energy is imparted to the molecules 
in contact with the metal and then to the adjacent water molecules, 
until all the water is heated. 

Heating by conduction only is in any case a relatively slow 
process; but, nevertheless, some materials conduct heat a great 
deal faster than others. Those substances that conduct heat 
best are called good conductors, while those that conduct heat 
poorly are called poor conductors or insulators. The metals 
are all classed as good conductors, while fluids (except mercury, 
which is a metal) are poor conductors. Silver is the best known 
conductor of heat and copper is next. All organic substances 
are poor conductors; this enables trees and other forms of vegeta- 
tion to withstand sudden changes in the weather without injury, 
since they heat slowly and cool with equal slowness. Moreover, 
the bark is a poorer conductor of heat than the wood beneath it, 
and this gives added protection. Rocks, earth, soot, and all 
loose materials transmit heat the more slowly as their density 
and uniformity of composition decrease. 

113. Coefficient of Conductivity. — The coefficient of conduc- 
tivity (also called the thermal conductivity) is the quantity of 



76 



ELEMENTS OF PHYSICS 



§4 



heat that will flow in one hour through a plate one foot thick, 
and having an area of one square foot when the difference of tem- 
perature between the two sides of the plate is one degree Fahrenheit. 
It may also be expressed in calories per second conducted by 
one cubic centimeter, temperatures being in degrees centigrade. 
Let ti and ^2 be the temperatures on the two sides of the plate, 
a = area of plate in square feet, h = thickness of the plate in 
feet, U = heat units in B.t.u. transmitted per hour, and k 
= coefficient of conductivity; then 

U = y(fx - t2) 

when ti is the higher temperature. 

The following table gives the value of k for a number of different 
materials; its value usually increases somewhat with the tempera- 
ture, for which reason, the temperature at which k was deter- 
mined is also given. When no temperature is given, it is under- 
stood to be 64°F. 

COEFFICIENTS OF CONDUCTIVITY 



Silver 244.0 

Tin 37.6 

Zinc 64.1 

Brass (63°)... 63.0 







Metals 




Aluminum 


.116.0 


Iron, wrought.. 


. .. .34.9 


Copper 


.220.0 


Steel 


. .. .26.2 


Gold 


.169.0 


Lead 


. .. .19.8 


Iron, cast (129°). 


.. 27.6 


Mercury (32°).. 
Fluids 


3.6 


Alcohol (77°)... . 


104 Water (63°) 


. 032 C; 



Benzine (41°)... .081 
Petroleum (55°) . 086 
Turpentine (55°)079 



Air (32°) 0126 

Ammonia (32°).. .0111 
Carbon Monoxide. 0121 

(32°) 



Carbon Dioxide 

(32°) 0079 

Hydrogen (32°) 0775 

Nitrogen (32°) 0085 

Oxygen (32°) 0136 



Insulating Materials 

Asbestos (100°) 097 Silk (100°) ... 028 Pulverized Cork (100°) . . 026 

Cotton (100°) 035 Wool (100°) . . 027 Infusorial Earth (100°) . 039 

Miscellaneous 

Cardboard 120 Linen 050 Rubber 100 

Cement 170 Mica 440 Sand 031 

Felt 022 Mineral Wool 035 Sawdust 037 

Firebrick 750 Paper 075 Vulcanite . .210 

Graphite 2.90 Porcelain 600 Wood Ashes 041 

114. An inspection of the table will show what poor conductors 
the fluids are as compared with the metals. Thus, the value of k 



§4 HEAT 77 

for steel is 26.2 -^ .0126 = 2000 times that for air, and the value of 
k for copper is 220 ^ .0126 = 17,460 times that for air. A layer 
of air is therefore a very good protection against loss of heat; 
hence, when making a boiler setting, it is always well to make the 
walls doable, with an air space between. Water is also a very 
poor conductor of heat; in fact, if a piece of ice be placed a httle 
below the top surface of a vessel of water and heat is apphed 
to the water above the ice, the water may be boiled without 
melting the ice. 

When steam is conveyed in pipes from point to point, the pipe 
should be covered with some good non-conducting heat material 
— asbestos, mineral wool, wood ashes, etc. — ^to keep the heat 
within the pipe. A covering of any kind, provided there is an air 
layer (even a thin one) between the covering and the pipe, will 
afford considerable protection against heat losses. Most pipe 
coverings owe their value to the air cells they contain. 

115. Convection. — If a kettle of water be placed over a hot 
fire, the water soon becomes heated and, later, boils. The water 
is heated all the way through, and the entire contents of the kettle 
have the same temperature, practically speaking. Very little of 
the heat imparted to the water is due to conduction, since if all 
the heat were due to this cause, it would take a very long time 
to raise all the water to the boihng point. What happens is this : 
The particles in contact with the heated surface are themselves 
heated, which causes them to expand; their density decreases 
owing to the increase in volume, and they rise to the top, colder 
and denser particles taking their place. There is thus created 
a circulation of water particles, the colder ones constantly taking 
the place of the warmer ones until they are all at practically the 
same temperature. This process is called convection. It is 
evident that convection can take place only in fluids, (hquids or 
gases). Without convection, the only way that a fluid could 
be heated throughout in a reasonable time would be by constantly 
stirring or otherwise agitating it. 

116. Radiation. — If one approaches a hot stove, or any hot 
body, a feehng of warmth is felt, which becomes the more intense 
the closer the body is brought to the source of heat. This feeling 
of warmth is not due to the heating of the surrounding air, 
because if a screen of some kind, a sheet, board, metal plate, 
etc., be placed between the body and the hot stove, the sensation 



78 ELEMENTS QE PHYSICS §4 

of warmth immediately disappears. Heat is thus transmitted 
or propagated through an intervening space from one object 
to another without heating appreciably the air between them. 
This method of transmitting heat is called radiation, and the heat 
thus transmitted is called radiant heat. All the heat received 
from the sun is radiant heat, the intervening space is about 
93,000,000 miles, and the question is, how is this heat transmitted? 
The answer to this question is a direct conclusion from the modern 
theory of heat. 

117. Dynamical Theory of Heat. — The modern dynamical 
theory of heat is embodied in the following statement from 
Ganot's Physics: 

"A hot body is one whose molecules are in a state of vibration. 
The higher the temperature of a body the more rapid are these 
vibrations, and a diminution in temperature is but a diminished 
rapidity of the vibrations of the molecules. The propagation of 
heat through a bar is due to a gradual communication of this 
vibratory motion from the heated part to the rest of the bar. A 
good conductor is one which readily takes up and transmits the 
vibratory motion from molecule to molecule, while a poor con- 
ductor is one which takes up and transmits the motion with 
difficulty. But even through the best of the conductors, the 
propagation of this motion is comparatively slow. How, then, 
can be explained the instantaneous perception of heat when a 
screen is removed from a fire or when a cloud drifts from the face 
of the sun? In this case, the heat passes from one body to an- 
other without affecting the temperature of the medium which 
transmits it. In order to explain these phenomena, it is imagined 
that all space, the space between the planets and the stars, as well 
as the interstices in the hardest crystals and the heaviest metal — 
in short, matter of any kind — is permeated by a medium having 
the properties of matter of infinite tenuity, called ether. The 
molecules of a heated body, being in a state of intensely rapid 
vibration, communicate their motion to the ether around them, 
throwing it into a system of waves which travel through space 
and pass from one body to another with the velocity of hght 
[about 186,400 miles per second]. When the undulations of the 
ether reach a given body, the motion is given up to the molecules 
of that body, which, in their turn, begin to vibrate; that is the 
body becomes heated, This process of this motion through the 



§4 



HEAT 



79 



ether is termed radiation, and what is called a ray of heat is 
merely one series of waves moving in a given direction." 

118. Laws Governing Radiation of Heat. — As the result of 
experiments, the following laws have been found to apply to heat 
radiation : 

The intensity of heat radiated from a given source varies (a) 
as the temperature of the source; (h) inversely as the square of 
the distance from the source; (c) grows less the greater the angle 
between the heat rays and a perpendicular to the surface heated. 

The first statement is to be expected. The second law follows 
from the fact that if S, Fig. 23 (a), be a source of heat, and ABCD 
and abed are parallel fiat 
surfaces whose perpendicu- 
lar distances from S are OS 
and oS, the solid formed by 
the rays drawn to the ver- 
tices is a pyramid, and 
from mensuration, area of 
abed : area of ABCD = oS^: 
OS^. But, the intensity of 
the heat on abed is greater 
than on ABCD, since its 
area is less; and the inten- 
sities on the two surfaces 
is inversely as their areas, 
which confirms the law. 
Regarding the third law, let S, Fig. 23 (6), be the source of 
heat and AB a, flat surface, SO being a perpendicular from S to 
AB. Evidently, any heat ray from S to AB that is not per- 
pendicular to AB is longer than OS; the greater the angle OS A 
or OSB the longer will be the Kne AS or BS, and the greater 
the area heated by the ray. Hence, the intensity of the heat 
at A or B will be less than at 0. 

Radiant heat is transmitted through a vacuum, because, 
according to theory, the vacuum contains ether, which transmits 
the vibratory motion, 

119. If two bodies having different temperatures are placed in a 
closed space, both bodies radiate heat energy; but the hotter body 
radiates more energy than the colder body, with the result that 
after a time both bodies will have the same temperature, since the 




80 ELEMENTS OF PHYSICS §4 

colder body will absorb more heat than the hotter body. Even 
after both have reached the same temperature, there will still be 
an interchange of heat energy between the two bodies. 

It is clear that the better conductor of heat a body is the more 
heat it will radiate. In the case of an uncovered steam pipe, the 
heat passes from the inner surface to the outer by conduction and 
then leaves the pipe by radiation. Therefore, unless this result 
is desired, as in a steam- or hot-water heating plant, the pipe 
should be covered with a non-conducting material, to keep the 
heat in the pipe. 

The amount of heat radiated depends upon the area of the 
surface radiating the heat and upon the material of which it is 
composed. Bright surfaces radiate more heat than dull ones and 
white surfaces more than black. Similarly, those surfaces that 
radiate the most heat absorb the least; it is for this reason that 
light colored clothes are worn in the summer and dark colored 
ones in the winter. Even if the clothes are made of the same 
material and have the same weight, dark colored clothes are 
warmer than hght colored ones. . 



EXAMPLES 

(1) How many heat units are equivalent to 1,980,000 foot-pounds of 
work? Ans. 2545 B.t.u., nearly. 

(2) How many calories are equivalent to (a) 2,655,229 foot-pounds of 
work? (6) to 1,980,000 foot-pounds of work? Ans. J 860. cal. 

\ 641.3 cal. 

(3) A piece of iron having a temperature of 1040°F. is placed in a copper 
vessel weighing 2 lb. 2 oz. and containing 12 lb. 4 oz. of water. The 
weight of the iron (wrought iron) is 1 lb. 6 oz. and the temperature of the 
water and vessel is 88°. Assuming that there is no loss by radiation, what 
is the temperature of the mixture? Ans. 101.16°F., nearly. 

(4) A number of steel rails are welded together until their length is 3000 
ft.; what will be the difference in length between summer and winter, if 
the range of temperature is from 110°F. to — 10°F.? Ans. 15.5 in., nearly. 

(5) An aluminum vessel when filled to a certain mark holds exactly 
1 U. S. gal. (2.31 cu. in.) at 62°F.; how much will it hold at 32°F.? 

Ans. 230.742 cu. in., nearly. 

(6) To find the temperature of the hot gases escaping from a blast fur- 
nace, a platinum ball weighing }-'2 lb. is placed in them; it is then placed 
in a brass vessel weighing 10 oz. and containing 3 lb. of water at a tempera- 
ture of 60°F. The temperature of the mixture being 72.4°F., what is the 
temperature of the gases? Ans. 2404°, say 2400°F. 



§4 HEAT 81 

CHANGE OF STATE 

120. Latent Heat of Fusion. — Suppose a piece of ice weighing 
1 pound and having the form of a right cyhnder with a circular 
base be placed in a cylindrical vessel of the same diameter; sup- 
pose further that the temperature of the ice is 32°F. and that 
the pressure of the atmosphere is 14.7 lb. per sq. in. If heat 
be applied to the vessel, the ice will gradually melt and become 
water. But, so long as there is any ice left, the temperature does 
not change; it remains at 32°, What becomes of the heat? 
The ice does not expand; in fact, after it is all melted, the volume 
occupied by the water is less than the original volume of the ice. 
According to the dynamical theory of heat, the effect of applying 
the heat to the ice was to overcome the attraction of cohesion 
to such an extent that the solid became a liquid. This heat is 
called latent heat; it cannot be measured with a thermometer. 
The heat that affects the thermometer is called sensible heat, 
because it is apparent to the senses. 

As the result of carefully conducted experiments, it has been 
found that it requires 144 B.t.u. of heat to change 1 pound of ice 
at 32° to water at 32°, the pressure being 14.7 lb. per sq. in. 
It has also been experimentally demonstrated that before water 
at 32° can be changed into ice, 144 B.t.u. of heat must be re- 
moved from every lb., the pressure being 14.7 lb. per sq. in. 
This value, 144 B.t.u., is called the latent heat of fusion of ice. 

The foregoing explains why water freezes (changes to ice) so 
slowly; it is necessary to remove a large amount of heat before the 
liquid can become a solid. It also explains why ice floats in the 
water in which it was formed; since the water expands in freezing, 
the density of the ice is less than water, thus causing the ice to 
float. Moreover, after the ice has become liquid (water), the 
water continues to contract in volume — become denser — until 
a temperature of 39° is reached; hence, the ice floats highest when 
the temperature of the water is 39°F. ; this temperature is called 
the temperature of maxinium density. 

The student will understand that heat values can be expressed 
in metric units (calories) by using grams, centimeters and degrees 
centigrade. The conversion of B.t.u. to calories, and vice versa, 
is explained in Art. 101. 

121. Latent Heat of Vaporization. — Assuming that the tem- 
perature is 39° F. and that the pressure is kept constant at 14.7 



82 ELEMENTS OF PHYSICS §4 

lb. per sq. in., further application of heat causes the tempera- 
ture of the water to rise and also causes the water to expand 
slightly, thus increasing its specific volume and decreasing its 
density. This action continues until a temperature of 212°F. is 
reached, when, if the pressure still remains the same, the water 
begins to vaporize, the temperature remaining 212° until all the 
water has been converted into vapor (steam). There is, how- 
ever, a great increase in volume. The heat required to change 
the water into steam of the same temperature and pressure is 
called the latent heat of vaporization, and when the pressure is 
14.7 lb. per sq. in., the temperature is 212°, and the number 
of heat units required is 970.4 B.t.u. per pound of water. When 
steam condenses and becomes water at this temperature, the 
same number of B.t.u. is given off or must be removed from the 
steam. Consequently, if 1 pound of ice and 5 pounds of water at 
32° are mixed with 1 pound of steam at 212°, the pressure being 
14.7 lb. per sq. in., the temperature of the mixture may be 
found approximately as follows: When the steam condenses, it 
gives up 970.4 heat units, which go to heat the water and ice, 
and becomes 1 pound of water at 212°. When the ice melts, it 
takes 144 heat units to change it into 1 pound of water at 32°, 
and this must be subtracted from the heat units in the mixture. 
The mixture thus consists of 1 + 5 = 6 pounds of water at 32°, 
1 pound of water at 212°, 970.4 B.t.u., and 144 negative 
B.t.u., the total weight of the mixture being 6+1 = 7 pounds. 

^^ , , . ,. , 6 X 32 + 1 X 212 + 970.4 - 144 
The temperature is therefore ^ 

= 175.8°. The result is approximate, because it has been assumed 
that the specific heat of water is constant and equal to 1 ; but, as 
before stated, the specific heat of water varies. However, the 
result as found is not far out of the way, and the illustration 
serves to show how the latent heats enter into problems of this 
Temperature State Specific nature, such as calculating the 
volume amount of steam required to 
1.0897 i"aise the temperature of liquids 
1.0909 in a mill. 

1.0001 122. The small table given 

1 0120 herewith affords some idea of 

1.0431 -l^ow the specific volume varies 

26.79 with the temperature. The vol- 

30.38 ume of 1 pound of water at 4°C. 



-10°C. 


= 14°F. 


Ice 


0° 


= 32° 


Ice 


0° 


= 32° 


Water 


4° 


= 39.2° 


Water 


50° 


= 122° 


Water 


100° 


= 212° 


Water 


100° 


= 212° 


Steam 


150° 


= 302° 


Steam 



§4 HEAT 83 

= 39.2°F. being taken as 1, the volume at 212° will be 1.0431 and 
the volume of 1 pound of steam at 212° will be 26.79 cu. ft. 
the pressure being constant and equal to 14.7 lb. per sq. in. For 
instance, the weight of a cubic foot of water at 39°F. is 62.4 
pounds, and the actual specific volume (volume of one pound) is 

^~ = .016026 cu. ft. Hence, 26.79 ^ .016026 = 1672 cu. ft. 

= the occupied by steam formed from 1 cu. ft. of water at 
39.2°F., the pressure of both water and steam being 14.7 lb. 
per sq. in. and the temperature of both 212°F. This shows the 
enormous expansion of water when converted to steam. 

123. Any vapor in contact with the liquid from which it is 
formed or having the same temperature and pressure that it had 
when it was formed is said to be saturated, if, in addition, it has 
no particles of the liquid entrained in it (mixed with it), it is said 
to be dry, and is then spoken of as dry and saturated. If heat 
be applied to a vapor that is dry and saturated, its temperature 
will increase; its volume will also increase, or if not allowed to 
increase, its pressure will increase, and the vapor is then said 
to be superheated. If superheated sufficiently, the vapor 
will exhibit all the characteristics of a perfect gas. 

If steam (or vapor) that is dry and saturated be subjected to 
any additional pressure, it immediately begins to condense; or, 
if the pressure is decreased, it is superheated. In other words, 
for any particular pressure, there is only one temperature for 
which the steam will be saturated. 

124. There is no simple formula showing the relation between 
the temperature and pressure of saturated steam, but the two 
following formulas, in which t = temperature and p = pressure, 
will give values accurate enough for all practical purposes : 

Between 10 and 250 lb. per sq. in. abs. (abs. means absolute), 

t = 241.5 - ^^ + 1.214p - .003385p2 + .00000448^3 (1) 

P 

For pressures between 1 lb. and 15 lb. per sq. in., abs., 

t = 121.42 - ^^^ + 11.892p - .5656p2 + .0126p3 (2) 
P 

Thus, for p = 10 lb. per sq. in. abs., formula (1) gives 

= 241.5 - ^^ + 1.214 X 10 - .003385 X 10^ 

+ .00000448 X lO'' = 193.2° 



84 ELEMENTS OF PHYSICS §4 

For p = 250 lb. per sq. in. abs., formula (1) gives 

t = 241.5 - ^^ + 1.214 X 250 - .003385 X 250^ 

+ .00000448 X 2503 = 401.1° 
The temperatures thus obtained are the ordinary tempera- 
tures as recorded on a Fahrenheit thermometer. 

125. What is called the total heat of steam is the number 
of heat units contained in one pound between the temperatures of 
32° and the temperature of the steam; it is equal to the number 
of heat units in the liquid, called the heat of the liquid, plus 
the latent heat of vaporization. Letting H = total heat, the 
following formulas will give the total heat accurately between 
p = 14.7 and p = 260 lb. per sq. in. abs., and may be used up 
to 300 lb. per sq. in. abs. : 

H = 1175.1 - ^^^^ + .17Sp - .000264p2 (i) 

For pressures less than 14.7 lb. per sq. in. abs., use the 
following formula: 

1 C 4.Q 

H = 1119.7 - i^^^ + S.2Sp - .07547?2 (2) 

For example, the total heat in one pound of dry and saturated 
steam when the pressure is 100 lb. per sq. in abs. is, by 

formula (1), 

405 Q 
H = 1175.1 - ^^ + .178 X 100 - .000264 X 100^ 

= 1186.2 B.t.u. 
This result, 1186 B.t.u. is the number of heat units that would be 
given up if 1 pound of dry and saturated steam under a pressure 
of 100 lb. per sq. in. abs. were condensed to water and the 
water were cooled to 32°; it is also the number of heat units that 
would be required to change 1 pound of water at 32° into 1 pound 
of dry and saturated steam at a pressure of 100 lb. per sq. 
in. abs. 

126. In Art. 124, the temperature of saturated steam when 
the pressure is 10 lb. per sq. in. was found to be 193. 2°F.; 
this is the boiling point of water for that temperature. In other 
words, when the pressure is 10 pounds per sq. in., the boiling point 
is 193.2° instead of 212°, the boiling point when the pressure is 
14.7 pounds per sq. in. The boiling point, therefore, depends 
upon the pressure, and this explains why liquids boil at a lower 



1 


101.8' 


2 


126.2 


3 


141.5 


4 


153.0 


5 


162.3 


6 


170.1 


7 


176.9 



§4 HEAT 85 

temperature in a partial vacuum. It also explains why a kettle 
of water boils aways so much faster when the barometer is low, 
just before a storm, than when it is normal, when the weather is 
fair. When the pressure is 100 lb. per sq. in., the boihng 
point of water, calculated as above, is 327.8°. 
Pressure Boiling The Httle table given in the margin shows 
abs. point the temperature at which water boils correspond- 
101 . 8° i^g ^^ the pressure in pounds per square inch given 
in the first column. For higher pressures up to 250 
lb. per sq. in., the boiling points may be found 
by substituting the pressure in the formulas 
(1) or (2) of Art. 124. It is to be noted that 
the temperature of saturated steam is the same 

8 182.9 as the boihng point of water for the same pres- 

9 188.3 sure; this follows from the definition of saturated 
10 193.2 steam. 

127. In the case of a partial vacuum, let i = inches of vacuum, 
as shown by the gauge; then the pressure in inches of mercury 
(the tension) is 29.921 — i. The temperature in degrees 
Fahrenheit and the pressure in pounds per square inch of dry 
and saturated steam in a partial vacuum may be calculated by 
the following formulas up to 28 inches of vacuum, letting 
w = 29.921 - i; 

t = 121.4 ~ + 5.84m - .1364^2 + .00149m3 (1) 

H = 1119.7 - ^^ + 1.611m - .01818m2 (2) 

m 

For example, if the vacuum gauge indicates 22 in., the tem- 
perature of the steam is, since m = 29.921 — 22 = 7.921, 

t = 121.4 - ;^^ + 5.84 X 7.921 - .1364 X 7.92P 

+ .00149 X 7.92P = 152°F, 
and the total heat of the steam is 

H = 1119.7 - 1^ + 1.611 X 7.921 - .01818 X 7.921^ 

= 1126.6 B.t.u. 

128. Melting and Boiling Points of Some Substance. — Every 
known substance can probably occupy all the three states of 
matter; if it is a gas, it can be liquified and then frozen (solidified) ; 
if a liquid, it can be frozen or vaporized; if a solid, it can be 
liquified (melted ) and then vaporized. It is, of course, understood 



86 ELEMENTS OF PHYSICS §4 

that all substances have not been made to occupy all three states; 
it is only recently that certain solids have been liquified, and not 
all of them have been vaporized. The following table gives the 

Fusion Boiling 

poiat point 

°F. °F. 

Alcohol, absolute (liquid) —170 173 

Ammonia (gas) — 104 —28 

Aluminum (solid) 1217 

Carbon (solid) 6300 

Copper (solid) 1980 4190 

Carbon Dioxide (gas) —110 

Gold (solid) 1944 

Helium (gas) —450 

Hydrogen (gas) —423 

Iridium (solid) 4172 

Iron (solid) 2740 

Lead (solid) 620 2777 

Mercury (liquid) -38 675 

Nitrogen (gas) —231 

Osmium (solid) 3992 

Oxygen (gas) '. — 182 

Platinum (solid) 3191 

Sea Water (liquid) 28 

Silver (soUd) 1760 3550 

Tantalum 5240 

Tin (solid) 450 4118 

Tungsten (solid) 5430 

Turpentine (liquid) 14 

Water, pure (liquid) 32 212 

Zinc (solid) . 786 1684 

fusion point or boiling point or both of a number of well-known 
substances, the ordinary state of each substance being given in 
parenthesis. It is to be noted that the fusion point is the tempera- 
ture (Fahrenheit) at which the substance changes from a sohd to a 
liquid (melts) or from a liquid to a solid (freezes) , and the boiling 
point is the temperature at which a liquid changes to a gas 
(vaporizes) or a gas changes to a liquid (liquefies) . In the case 
of carbon, there is apparently no liquid state, it changing directly 
from the solid to a gas, but the gas has never been isolated and 
examined, the temperature being too high. 

129. Humidity. — There is always a certain amount of water 
vapor present in atmospheric air, and the pressure of this vapor 
is usually different from that of the air. For any particular tern- 



§4 HEAT 87 

perature of the air, there is a pressure for which the water vapor 
will be saturated; that is, any increase in pressure or any decrease 
in temperature (however small) will cause the vapor to condense. 
In general, the water vapor is in a superheated state, its pressure 
being less than that required for saturation. Let w be the weight 
of the vapor per cubic foot when the temperature of the air is t, 
and let w' be the weight per cubic foot at the same temperature 
when the vapor is saturated; then w' will be greater than w, and 

w 
the ratio — is called the relative humidity, which has different 

values for different temperatures and pressures. 

Suppose that for some particular state of the atmosphere the 
temperature is t and the barometric pressure is 6, and let w be the 
weight per cubic foot of the water vapor. If the air be cooled, 
both the temperature and pressure will fall; the water vapor also 
cools and after a time, it will reach its saturation temperature f and 
saturation pressure b'; this temperature is called the dew point 
corresponding to the temperature t and pressure b. Hot air absorbs a 
greater amount of moisture than cold air, which explains the heavy 
dews of summer. During the night, the air cools until the dew 
point is reached, and any further cooling causes the water vapor 
to condense and fall as dew. 

130. The relation of humidity to health and comfort is of 
extreme importance. Hot, damp days are oppressive and seem 
hotter than they really are because the relative humidity is so 
high, the vapor is so near the dew point, that free evaporation of 
moisture from the body is interfered with; on the other hand, 
when the relative humidity is low, the air is dry and the evapora- 
tion of moisture from the body is too great. In houses and 
offices, where the temperature is kept at about 68° or 70°, the 
best results are obtained when the relative humidity is from 
50% to 60%; if the relative humidity is less than this, it will 
require a higher temperature for comfort on cold days; and if it is 
much less, the result will be apparent by the widening of cracks 
in floors (if of wood), the loosening of the furniture, etc., and it 
will also have a bad effect on the health. 

In drying paper and pulp, the relative humidity in the dry 
loft or machine room is an important factor, since the rate at 
which the air takes moisture from the stock becomes slower as 
the moisture already in the air increases; on the other hand, the 
amount of moisture in stored paper varies with the moisture 



88 ELEMENTS OF PHYSICS §4 

in the air, and the quahty is Hkewise affected by this factor. 
A well insulated roof and an interior temperature above the dew 
point of the moisture-laden air is required to prevent condensation 
in the loft or machine room. 

131. Foaming and Surface Tension. — If a fine needle be placed 
on the free surface of a liquid, care being taken to have its axis 
parallel to the plane of the free surface and to lay it gently on the 
surface, the needle will float, and this despite the fact that the 
density of the needle is from to 6 to 8 times that of the liquid. 
Small, dust-like particles of any kind, though many times as 
dense as the liquid, will float provided they come into contact 
with the liquid without the exertion of a downward force that is 
greater than that due to gravity. But if the needle or the particles 
are at any time entirely submerged, they will sink, if of greater 
density than the liquid. This phenomenon is due to what is 
called surface tension, which is very closely related to capillarity, 
and is explained as follows: 

The molecules (and particles) of a liquid attract one another, 
with the result that they tend to form compact masses; this 
attraction takes place in all directions, and when the masses are 
small, they tend to form little spheres, thus reducing the free 
surface to a minimum, a sphere having the smallest surface for a 
given volume. Thus, if a match be dipped in a liquid and then 
held up so the liquid can run down to one end, a round drop will 
form, and if the drop falls, it has the shape of a perfect sphere. 
The size of the drop will depend upon the cross-sectional area of the 
end of the match or stick and upon the viscosity of the liquid. 
Viscosity may be defined as the state of fluidity; it may also be 
thought of as internal friction. For instance, if a thin, flat 
plate be drawn through a liquid in a direction parallel to the plane 
of its flat surface, the force required to pull the plate is a measure 
of the viscosity of the liquid; it will evidently require a greater 
force to pull the plate through molasses than through water; 
hence, molasses is more viscous — has greater viscosity — than 
water. 

Now when the upper surface of a liquid is free, there is no 
attraction from above, but the molecules in the immediate neigh- 
borhood below the surface of the liquid tend to pull downward. 
There are also other forces pulling the surface molecules and 
tending to bring them as close together as possible; that is, so that 
the area of the exposed surface will be as small as possible. This 



§4 HEAT 89 

causes the free surface to behave as though an elastic skin were 
drawn tightly over it, and it creates the surface tension mentioned 
above. 

132. When a liquid is violently agitated, as by rapid stirring, by 
the use of an egg beater, by heating, etc., bubbles form on the free 
surface; this phenomenon is called foaming. The bubbles are due 
to surface tension, which enables them to keep their form. The 
amount and quality of the foam that forms on any liquid depends 
upon its composition, viscosity, and temperature. The presence 
of impurities will freqently cause the liquid to foam more readily 
than when pure. When pure water is boiled, bubbles form on the 
free surface, but they do not last, collapsing almost as soon as 
they form; the same thing happens when water is violently 
agitated with an egg beater. If, however, the water is not pure, 
but contains other substances in solution, making it more viscous, 
the bubbles will be larger and will retain their form longer; in 
fact, they may retain their form until they collapse from eva- 
poration as in the case of soap suds, etc. Such bubbles are com- 
monly called froth. 

133. In paper mills, the froth on top of a liquid is sometimes 
but the dried mechanical structure of evaporated films, which, 
upon examination, will be found to be superimposed layers of 
more or less dried clay, rosin, calcium resinate, fibers and other 
solids, all of which act to form a structure for fresh bubbles to 
adhere to and prevent the bubbles proper from breaking. If this 
structure is allowed to accumulate, it may sink, mix with the pulp 
mixture or the paper stock, and show up as dirty spots in the 
product. A spray of water is sometimes effective in breaking up 
this structure, but it is advisable in most cases to skim it off and 
then endeavor to prevent its formation by one or more of the 
following methods: 

(a) By spraying the surface (if it has a tendency to foam) with 
a fine needle spray of pure or clean water, thus breaking the 
bubbles before they attain any great size, 

(6) By adding some solution to the liquid that will increase its 
surface tension, thus preventing it from breaking up into air 
bubbles. 

(c) By adding clean water to the mixture. 

(d) By lowering the temperature of the liquid and increasing 
the temperature of the air in the neighborhood of the liquid. 



90 ELEMENTS OF PHYSICS §4 

(e) By mechanical removal of the froth as soon as formed by 
the use of a skimming device. 

In actual practice, it has been found in some mills that when 
the temperature of the water used on the screens is lower than 
25°C. (77° F.), there is less foaming than when the water is above 
this temperature. 

134. In the case of steam boilers, what is called priming occurs 
when water bubbles are carried over with the steam. Consider- 
able water may thus be added to the steam, thus making it no 
longer dry, and serious results may follow, A relatively large 
amount of water may thus accumulate in the cylinder of the engine 
or turbine, and the swiftly moving piston or blades may strike 
this, causing the cylinder to burst, since water is practically 
incompressible. Priming is usually due to impurities, which may 
be contained in the feed water or may be added to it by too 
liberal use of boiler compounds, added to prevent incrustation. 
When feed water that is known to foam must be used, an analysis 
should be made of the water to determine what impurities are 
present; it is then frequently possible to add some chemical that 
will neutralize these impurities before the water is fed into the 
boilers. Priming may sometimes be overcome by changing the 
water frequently in the boilers, by blowing them down and filling 
them up with fresh water, or by adding some water glass (sodium 
silicate), which increases the surface tension. 



LIGHT 



RADIANT ENERGY 

135. Nature of Light. — An explanation of the dynamical 
theory of heat was given in Art. 117, and it was there stated that 
heat was propagated by reason of vibratory motions set up in the 
ether, which pervades all space and all bodies. This vibratory 
motion results in a series of waves, somewhat like those which are 
formed when a stone is thrown into a pond when the water is still. 
The particles of water move up and down, communicate their 
movement to the adjacent particles, and the wave moves outward 
from where the stone entered the water. Note that it is the 
particles of water that move up and down and that it is the wave 
form that moves outward; the water particles all return to the 




§4 LIGHT 91 

place from which they started. This is practically what takes 

place in the ether, the ether particles moving transversely to 

(across) the direction of the wave movement. If a plane section 

be taken through a wave of this kind, the result will be something 

hke the curve in Fig. 24, in which the parts AaB = BbC = CcD 

= etc. and AaB is symmetrical to BbC with respect to the point 

B, BbC is symmetrical to CcD 

with respect to the point C, K~r 

etc. The distance between a /^ 

any two corresponding points 

on two symmetrical parts of 

the curve, as AC = ac = hd p^^ 24 

= etc. is called a wave length, 

here designated by L The highest points, b, d, etc., are called 

crests, and the lowest points c, e, etc., are called troughs. The 

perpendicular distance between a crest and a trough, here 

designated by h, is called the amplitude of the wave. 

136. All radiant energy is propagated (transmitted) in the 
form of waves. Heat and light are both forms of radiant energy, 
the sole difference between them being in the length of the waves; 
if the waves are too long, they produce only the sensation of heat, 
and if they are too short, produce neither the sensation of heat 
nor light, but will produce chemical effects and are then called 
actinic rays. Actinic rays will affect the photographic plate. 

137. Bodies that act as a source of light, that is, which are able 
to produce vibrations in the ether of such a nature as to make 
them visible, are called luminous bodies; as a piece of red hot 
iron, a candle flame, an incandescent electric lamp, etc. Such 
bodies are said to emit light. Some luminous bodies can be seen 
to emit hght only in the dark or in semi-darkness; this is because 
the light of day is so much stronger that it overpowers the light 
from the luminous body; thus, the hght of a lightning bug, of a 
piece of wet phosphorous, the light of the stars, etc. cannot be 
perceived in full daylight. Bodies that do not emit light and 
can be seen only by light that is reflected from a luminous body 
are called non-luminous. Most bodies are non-luminous under 
ordinary conditions. 

A ray of light is the line along which light is propagated; rays 
issue from the source of light in all directions, and if the medium 
through which they pass is uniform in its structure, every ray of 



92 ELEMENTS OF PHYSICS §4 

light is a straight line. A collection of parallel rays make up a 
beam of light; but if the rays are not parallel, either converging 
toward or diverging from a point, they form a cone or pencil. 

When light passes through any substance whatever, a solid, a 
liquid, or a gas, the substance is called a medium. If light rays 
pass freely through the medium, so that objects are clearly seen 
when looking through it, the medium is transparent; clear water, 
air, ordinary window glass, glassine paper, certain crystals, etc. 
are transparent. If objects are only faintly seen through the 
medium, it is said to be semi-transparent; smoked glass, a 
foggy atmosphere, slightly muddy water, etc. are semi-transpar- 
ent. If light can be seen through the substance, but the form 
of objects cannot be distinguished, the medium is called trans- 
lucent; ground glass, colored glass, certain crystals, etc. are 
translucent. Substances which allow no light to pass through 
them are called opaque; a piece of iron, a stone, a stick of wood, 
heavy cardboard, etc. are opaque mediums. However, if 
any substance be cut in sufficiently thin slices or rolled suffi- 
ciently thin, it will be more or less translucent. Opacity is a 
desirable quality in printing papers. 

When the medium is transparent, all the light that strikes it 
goes through it; but when the medium is translucent, some of the 
rays are absorbed and the remainder go through, those that are 
absorbed being transformed into heat. Thus, a clean, plate- 
glass window exposed to the sun in the summer will be cool; but 
if it be covered with paint or varnish or the surface be roughened, 
some of the light rays will be absorbed and the glass will become 
warm — it may even become hot. In the case of opaque sub- 
stances, all the light rays striking it are transformed into heat. 

138. Light Rays are Right Lines. — It was stated in the last 
article that every ray of light is a right line, provided the medium 
through which it passes is uniform; thus, a ray of light going 
through a clear atmosphere, clear water, glass, etc. will be straight ; 
but, as will be explained later, it will not be straight throughout 
its length, if it passes through two or more different mediums. 
That the rays are right lines is easily proved by the fact that a 
person can see through a straight tube, but cannot see through 
it if it is bent. So long as the medium is uniform the rays are 
straight ; but if a ray passes through several mediums, it will be a 
broken line, that part of the ray through each medium being 
straight. A ray can never be a curved line. 



§4 



LIGHT 



93 



139. Velocity of Light. — The speed of hght is so great that it is 
difficult to measure it with any great degree of exactness. How- 
ever, the velocity has been determined in a number of different 
ways, the most probable value being 186,400 miles per second. 
This is equivalent to 300,000 kilometers = 3 X lO^'^ centimeters 
per second, the latter value being very easy to remember. An 
object moving at this speed would be able to go around the 
earth about 7>^ times in one second. The distance from the 
earth to the sun is about 93,000,000 miles; consequently, hght 
received from the sun requires 93000000 h- 186400 = 499 seconds 
= 8 min. 19 sec. to reach the earth; in other words, the sun has 
risen above the horizon 8 min. 19 sec. before it is seen, and it is 
seen for 8 min. 19 sec. after it has set. Anything that happens on 
the sun cannot be seen until 8 min. 19 sec. after it has occurred. 
The fixed stars are so far away that it takes over 3 years to reach 
the earth from the nearest one, and some are so distant that it 
takes light thousands of years to reach the earth. 



SHADOWS 

140. Shadows Cast by a Luminous Point. — If a luminous 
body be very small, so that for practical purposes it may be 
considered to be a point, it is called a radiant. Hence, any 




Fig. 25. 

luminous body may be considered as a collection of radiants 
distributed over its surface. In Fig. 25, let P be a radiant and ^ 
a flat, opaque, white screen having a smooth dull surface that will 
not reflect light. Let AB be a flat, opaque plate of irregular 
outhne, situated between P and S. Rays of hght from P shoot 
out in straight hnes in all directions— upwards, downwards, 



94 



ELEMENTS OF PHYSICS 



§4 



sideways, forwards, and backwards — and illuminate the screen S. 
The rays, however, cannot pass through the plate AB, and the 
result is that there is a dark spot CD on the screen, which is called 
the shadow of AB. The outline of the shadow is of the same form 
as that of the object AB; it is really a projection of AB on the 
screen formed by drawing right lines from the point P, touching 
the perimeter of AB and intersecting the surface of the screen. 
These lines form a pencil (Art. 137) of rays, and any ray included 
within the bounding surface of this pencil is stopped by the ob- 
ject and does not strike the screen. This is another proof of the 
fact that the rays of light are straight. The outline of a shadow 
formed by light proceeding from a single radiant is sharp and 
distinct. 

141. Shadows Cast by a Luminous Body. — If the shadow is 
cast by a luminous body instead of a single radiant, a somewhat 
different result is obtained. Referring to Fig. 26, suppose XF to 



r 


i 

h 


E^^^^^~-^^ 


____^ 


R ^ 


s 
w 


^ 






A 


^~~~~~~~--~JL---' 


^^ 


-— — _ T 


i 


/ 


,.------'' — ~~~~~~~-^V MS 


T^ 




V 


^ 


F ^^^.^-^ 


^ 


Z> 




] 


r 








JV 





Fig. 26. 



be a luminous sphere, RM an opaque sphere, and let SN be the 
edge of a screen; then, if XR and YM are tangents to the two 
spheres, all radiants on the surface XA Y emit rays of light that 
touch the object RM; but all other radiants on the sphere XY 
emit rays that do not touch RM. Considering the radiant X, 
it forms a pencil RXM, which results in the shadow WD on 
the screen; the radiant Y forms the pencil MYR, which results 
in the shadow VC; the total shadow cast by these two radiants 
is CD. But, the part WC is illuminated with more or less inten- 
sity by radiants situated between X and F, the shadow becoming 
lighter and hghter as the distance from W toward C increases, 
since rays from a greater number of radiants strike the part WC 
of the screen above the line ER than below it, radiants situated 
between E and F also striking this part of the screen. The same 



§4 LIGHT 95 

thing occurs in connection with the part VD of the shadow, and 
as a consequence, the edges C and D are very faint and indistinct. 
The part between W and V is not illumined by rays from any 
radiant; this part is equally dark throughout and is called the 
umbra. The part outside of the umbra, represented by WC and 
VD graduates from the color of the umbra to full illumination, 
and is called the penumbra. The case just described fulfills all 
the conditions of a total eclipse of the sun. Here XF is the sun, 
RM is the moon, and the screen SN is the earth. Persons in the 
umbra witness a total eclipse; those in the penumbra, see only 
a partial eclipse. 

142. Brightness and Intensity. — The brightness of a Hght does 
not depend upon the size or shape of the luminous body, just 
as the density of a substance does not depend upon the size or 
shape of the body. A pin point of light may be just as bright as a 
powerful search light. The brighter the hght the darker will 
be the shadow it casts, since the illuminated part around the 
shadow will then be brighter and there will be a greater contrast 
between the shadow and the illuminated portion of the screen. 
The intensity of a light however, depends upon the brightness 
of the radiants, their number (and, consequently, the area of the 
surface emitting the rays), and the distance of the object illu- 
minated from the source of light. In fact, the intensity of light, 
like the intensity of heat or any other form of radiant energy, 
varies inversely as the square of the distance. The proof of this 
is exactly the same as was given in the case of heat. Art. 118. 

143. Candlepower. — The common standard for the intensity 
of light is one candlepower, which is the amount of light emitted 
by a sperm candle % inch in diameter when burning at the rate 
of 120 grains per hour. This is called a standard candle. If, 
therefore, a standard candle illuminates a certain object 3 ft. 
distant with the same intensity as another source of light situated 

• 12 feet distant, then, the candlepower of the standard candle being 

1, let c be the candlepower of the other light. According to the 

the law of the last article, c may be found from the proportion 

144 
c : 1 = 12^ : 3^, or c = -q- = 16 candlepower. 

It is to be remembered that the proportion is an inverse one. 

144. Photometers. — A photometer is an apparatus for meas- 
uring the relative intensities of light. Photometers are made in 



96 



ELEMENTS OF PHYSICS 



§4 



various forms, two of which will be described here, as they make 
use of two different principles. 

(a) Rumford's Method. — Referring to Fig. 27, T is a horizon- 
tal, flat surface on which is erected a vertical, flat, white screen S. 
R is an opaque rod, C is a standard candle or other source of 
light whose intensity is known, and L is a light whose intensity 
or relative intensity as compared with C is to be measured, say 
an incandescent electric lamp. C casts a shadow of the stick or 
rod R at a, which is illuminated by the light from L, and L casts 




Fig. 27. 

a shadow of the rod R at b, which is illuminated by the light from 
the candle C. By moving C back and forth along the line Ca, 
it will be found that there is one point for which the two shadows 
will be of the same distinctness ; suppose this to be the point C in 
the figure. Measure the distances Ch = m and La = n; then, if 
the intensity of the candle be represented by C and of the lamp 
by L, 



L:C = 



or L 






If C is 1 candle power, then L = — ^ candlepower. In other 

words, when two sources of light illuminate an object with the same 
intensity, these candle-powers are directly proportional to the squares 
of their distances from the object. 

(b) Bunsen's Method. — The essential part of a Bunsen 
photometer is a disc of unglazed paper having a round grease spot 
in the center, thus making the paper translucent at this spot. 
The disc B, Fig. 28, is held in an adjustable holder, which can be 
moved along a graduated scale S. A standard candle C or other 
source of illumination whose intensity is known is placed on one 
side of the disc, and an electric lamp L or other light whose inten- 
sity it is desired to find is placed on the other side. If the lamp 



§4 



LIGHT 



97 



be turned out and the candle be left burning, the side of the grease 
spot nearest the candle will be lighter than the other side. If, 
therefore, both lights are going, that side of the grease spot will 
be the darker which is turned toward the light of less intensity. 
By moving the holder B along the scale, a point will be found at 




ifi'i I in Ti 




I I I I ff l I I I ' f Vl' q I 1 1 1 1 1 1 I 1 1 I 1 1 1 1 1 1 I I I I I 1 1 1 1 I I I 1 1 I I I I 1 1 1 1 1 1 M I I III I I 1 1 1 1 1 1 I I I 1 1 M I I I I I I I 11 1 I III I f 

10 20 30 40 SO * 60 70 80 90 



Fig. 28. 



which both sides of the spot are illuminated equally. Then 
measuring the distances m and n between the disc and C and L, 
respectively. 






the same result as was obtained by Rumfords' method. 



REFLECTION OF LIGHT 

145. Reflection from Plane Surfaces. — Let P, Fig. 29, be a 
polished plane surface, as a mirror, polished steel, etc., and let C 
be a radiant. Suppose a ray of light from C to strike the surface 
P at .A ; instead of being absorbed, the light will be reflected at 
A along the line AE, and if the eye is situated at G, the reflected 
ray will enter it. Pass a plane through A and C perpendicular 
to plane P; it will be found that the reflected ray AE Hes in this 
plane, and a perpendicular P'A at A also hes in this plane. The 
angle CAP' is called the angle of incidence, and the angle P'AE 
is called the angle of reflection. It has been proved by repeated 
experiments and measurements that the angle of incidence is 
always equal to the angle of reflection, that is, CAP' = P'AE and 
P'A bisects the angle CAE. Hence, to draw a reflected ray 
from a plane surface, draw a perpendicular to the surface at the 



98 



ELEMENTS OF PHYSICS 



point where the incident ray strikes it, the point A in Fig. 29; 
then from A draw AE so that angle P'AE = P'AC, and AE will 
be the direction of the reflected ray. For any other ray, as CB, 
draw a perpendicular at B, and then draw BF so that the angle 
included between BF and the perpendicular BP" will equal the 
angle included between BC and the perpendicular. 




Fig. 29. 

The line XY passing through the point A is the trace of the 
plane that includes AC and AE, and if CD be drawn perpendicular 
to XY it will lie in this perpendicular plane, since it will then be 
parallel to P'A. Produce EA until it intersects CD in D; then, 
from geometry, angle YAE = angle XAD, and since YAE 
= XAC, XAD = XAC, the triangle CAD is isosecles, and CA 
= DA. With the eye in the position shown, the point C will 
appear to be located at D; that is, the apparent position of the 
radiant, as viewed by the eye, will appear to be as far below the 
plane P as it actually lies above the plane P; and this will be the 
case no matter where the eye is situated above the plane of re- 
flection, since the triangle formed by CD and the sides drawn from 
C and D will always be isosceles and the length of the base CD is 
not changed. This will still be true, even when the eye is di- 
rectly over C and the triangle has become a right line of which 



§4 LIGHT 99 

CD is part. This effect will be apparent in the reflections pro- 
duced in any plane mirror; the object reflected will always appear 
as far back of the mirror as it actually is in front of it. 
146. Reflection from. Curved Surfaces. — The law of reflection 
for curved surfaces is the same as for plane surfaces, provided 
the angle of incidence and reflection are those included between 
the incident and reflected rays and the normal to the surface at 
the point of their intersection. A normal to a surface at any 
point is a perpendicular drawn to a tangent to the surface at that 




^c" 



point. For instance, a radius is always normal to a circle or 
sphere at the point where it intersects the circle or sphere, 
because a tangent drawn at that point is perpendicular to the 
radius. For other curved surfaces, special geometrical construc- 
tions are required to draw a tangent at any particular point; but 
when a tangent has once been drawn, a perpendicular to the tan- 
gent at the point of tangency will be normal at that point. 
Having drawn a normal to the surface at the point where the inci- 
dent ray strikes it, measure the angle between the incident ray 
and the normal, pass a plane through the lines defining the inci- 
dent ray and the normal, and lay off in this plane on the other side 
of the normal an angle equal to the one measured ; the side of this 
angle will have the same direction as the reflected ray and will coin- 



100 ELEMENTS OF PHYSICS §4 

cide with it. Thus, suppose MN, Fig. 30, to be a section of a 
spherical surface whose center is 0; let C be a radiant and C'A' a 
ray. Draw OA' and it will be a normal to the surface at A' ; draw 
A'B' so it will make an angle B'A'O equal to the angle C'A'O, 
and A'B' will be the reflected ray when C'A', OA', and B'A' all 
lie in the same plane. Again, if C" be a radiant and C"A" a 
ray, draw OA'" through and A"; it will be a normal to the 
surface at the point A". Lay off A"B" so angle A"'A"B" 
= A"'A"C", and A"B" will be the reflected ray. The arrow- 
heads indicate the direction of the incident and reflected rays. 
If that part of the curved surfaces facing toward A'" be pol- 
ished, thus reflecting all the rays striking it, and the eye be 
placed anywhere along A"B" , the radiant C" will appear as 
though located at D. To find the point D, draw TT' tangent 
to the surface at A" and in the same plane as A"B" and A"C"; 
draw B"C" parallel to TT' and C"D parallel to OA'"; pro- 
duce B"A" until it intersects CD in T), which is the required 
point. 

147. Diffusion of Light. — Suppose a beam of light to strike a 
smooth, polished plane surface, say a plane mirror, as shown at 
(a), Fig. 31. The rays will all be reflected in parallel lines and 
will illuminate any object that they strike only to the extent 
of an area equal to the area of a cross-section of the beam. 





Fig. 31. 

Now suppose the reflecting surface to be comparatively rough, 
like ground glass, unglazed white paper, etc. The incident rays 
will all be parallel as before, but the reflected rays will take 
practically every direction, as indicated in (6), with the result 
that every part of the object will be illuminated and, in addition, 
all parts of the room as well. The light is then said to be diffused ; 
it will illuminate no part of the room with the same intensity as in 
(a), but it will be softer, less trying on the eyes, and will illuminate 
a far greater area. The manner in which book paper reflects 
light is an important consideration. Very smooth, glazed paper 
is trying to the eyes, especially under artificial light. 



§4 LIGHT 101 

148. Visibility of Objects.^ — ^In the case of a self-luminous 
body, the Hght rays are transmitted directly to the eye, and the 
form of the body is sharply outlined. When a body is non- 
luminous, however, as is usually the case with visible objects, 
it can be seen only by reflected light, which is almost invariably 
diffused, the result being that a non-luminous body is not usually 
as distinct as a luminous one. The diffusing surface may be 
considered as made up of an extremely large number of elementary 
areas, each of which reflects light in all directions from a lumi- 
nous source, thus making visible the outline of the body, no 
matter where the eye is placed within the limits of visibility. 
A perfectly reflecting surface or a transparent one is not visible 
to the eye. The water in the gauge glass of a steam boiler cannot 
be seen when clear; if the gauge glass is only partly full, the 
steam above it cannot be seen; and the only reason that the 
surface of the water in the glass can be seen is because of the 
difference in refractive powers of water and the air or steam above 
it. A polished reflecting surface, like a plate-glass mirror, is not 
visible, since the reflected rays are the same in all respects as the 
incident rays, and only the luminous source is seen. This is 
proved by the fact that if the wall of a room were made one laige 
plane mirror, a person walking toward it will not recognize the 
wall until he strikes it. A black body does not reflect rays to the 
eye, but is made visible by rays reflected from its surroundings. 



REFRACTION OF LIGHT 

149. Index of Refraction.^ — ^In Fig. 32, H'H represents the 
water level of a body of water, and AO is a light ray that strikes 
the water (a plane surface) at 0. A part of the light is reflected 
along the line OA'and the rest enters the water. Now instead 
of continuing through the water in the direction of the incident 
ray AO and along OB', the ray is deflected at and takes the 
direction OB. The angle B'OB is called the angle of deviation, 
and the change in direction of the incident ray when passing 
through the water is called refraction; AO h called the incident 
ray and OB the refracted ray. Through 0, draw C'C perpendicu- 
lar to H'H; then AOC is the angle of incidence and BOC 
is called the angle of refraction. It is to be noted that the incident 
ray and the refracted ray are on opposite sides of the perpendicular 
drawn at the point of incidence 0; also, the angle of refraction 



102 



ELEMENTS OF PHYSICS 



§4 



increases as the angle of incidence increases, and vice versa; 
hence, when the incident ray is perpendicular to the surface H'H, 
there is no refraction, both the incident ray and the refracted ray 
being normal to H'H; but for any other position, a ray is always 
refracted when it enters or leaves a different medium. With but 
few exceptions, when a ray passes from a transparent medium of 
less density (as air) through or into one of greater density (as 
water, glass, etc.), the refracted ray is inclined towards the normal 
drawn through the point of incidence; thus, in Fig. 32, OB is 



A" 
A \ 


n 
J 

n / 




rr 0\ 


g^^g-g 


^-=-iE-^^^E2rE3^:^^^:i=:=^^:r^:^^- 


^fcllfl 


^^ 


E:=i>^:i=--:>^i>:>i:-e-d^ 


r-::-z-^5^:^:irrF:i^:5"^;=^=i:' 




., B B' 





Fig. 32. 

inclined towards OC, and BOC is less thanAOC. Consequently, 
also, when a ray immerges from a medium of greater density into 
one of less density, the refracted ray is inclined away from the 
normal; thus, in Fig. 32, if BO be the incident ray, it will not 
follow the dotted line OA" (which is BO produced), but will take 
the direction OA, being inclined farther from the normal C'C, 
and AOC is greater than A'VC = BOC 

At any point m on OA, draw a perpendicular mn to OC; 
then mnO is a right triangle, right-angled at n, and Om is the 
hypotenuse. The side mn is called the side opposite the angle 0. 
Now, in any right triangle, the ratio of the side opposite an angle 
to the hypotenuse is called the sine of that angle; this may be 
expressed as 

side opposite 



sine = 



hypotenuse 



mn 



For example, in the right triangle mnO, sine = -p^—, sine m = 

Om 



On 

Om' 



§4 LIGHT 103 

and sine n = -J^ = 1, that is, the sine of 90° (a right angle) is 1. 

Here sine 0, sine m, etc. mean sine of angle 0, sine of angle m, etc. 
If from the point p, a perpendicular be drawn to C'C, pqO will 
be a right triangle, right-angled at q, and pO will be the hypotenuse ; 

then sine pOq = jy- It has been found by experiment and 

observation that the ratio of the sines of the angles of incidence to 
the corresponding angles of refraction is constant; that is, no 
matter what the inchnation of the incident ray may be, the ratio 
of the sine of the angle of incidence to the sine of the angle of 
refraction always has the same value, which is called the index 
of refraction. In Fig. 32, 

sine AOC (angle of incidence) __rrm^pq^ ^ mn X Op ^ mn 

sine BOC (angle of refraction) ~ Om ' Op ~ pq X Om pq 
when Op is made equal to Om, as is usually done. 

The index of refraction varies for different mediums ; in tables 
that give the values of the index for various mediums, the ray 
of light is supposed to pass from a vacuum into the given medium, 
and the indexes of refraction are called absolute indexes. It 
may be remarked that the absolute index of refraction for air is 
so nearly 1 that it may be considered as 1 in practice; in other 
words, there is practically no refraction in air. 

150. If the ray pass from a denser medium into a rarer one, as 
from water into air, the refracted or emergent ray is inclined 
away from the normal. Thus, referring to Fig. 33, let AO be a 
ray that leaves the water at 0; it is refracted and has the direc- 
tion OB in the air. If the ray had the direction CO, the refracted 
ray would have the direction OD, and would lie in the upper 
surface of the water. The angle COP' for which the emergent 
ray lies in the surface of contact between the two mediums is 
called the critical angle; for water, its value is 48° 33', say 48>^°. 
If the incident ray make a still greater angle of incidence, it will 
not leave the water at all, but will be reflected back in the same 
manner as though the surface of the water were a mirror. This 
is called total reflection, because no part of the ray is absorbed by 
the reflecting surface; thus, the ray EO is reflected along OF. 
This phenomenon appHes to other transparent substances, as 
glass and quartz, and is made use of in the construction of some 
optical instruments. A prism face can thus be made to reflect 
certain rays and to transmit others. 



104 



ELEMENTS OF PHYSICS 



§4 



The direction of a ray of light in a medium of uniform density- 
is straight, that is, it is a right hne. If a ray pass from one me- 
dium, as air, through another, as glass, and on into the first 
again, the ray wiU be refracted twice and the incident and emer- 
gent rays will be parallel if the surfaces of the refracting sub- 
stance be parallel. 

It is because of refraction, that a straight stick or rod appears 
bent when partly submerged in water. For instance, the stick 
S, Fig, 33, will appear as fgh instead of as fga when the eye is 




Fig. 33. 

situated at e. If the entire stick were in air, a ray from a to the 
eye would lie along ae; but on account of the refraction, the ray 
is bent, and takes the direction he, and does not enter the eye. 
The ray from a that enters the eye has the direction ad in the 
water and de in the air; this makes the end of the stick appear 
to be at h, h being on the line ed produced. As h is nearer the 
surface than a, the depth of any object below the surface does 
not seem as great as it really is (unless the eye is in a vertical line 
over the object. A small fish lying in the water at a will appear 
to be at h when the eye is anywhere along the line ae. 

When a ray of light is capable of being refracted, it is said to be 
refrangible, the word refrangible being used instead of refract- 
able. As will presently be shown, certain kinds of light rays are 
more refrangible than others. 

151. Refraction through a Prism. — ^Let ABC, Fig. 34, be a 
cross-section of a transparent colorless glass prism, and let M 
be a radiant. Draw CD bisecting the angle C; suppose the prism 



§4 



LIGHT 



105 



to be so located that CD is vertical. It will evidently be possible 
to place the prism in such a position, with CD vertical, that a ray 
MN will be refracted along NP, perpendicular to CD; NP will 
then be horizontal. When the refracted ray emerges from the 
prism at P, it will take the direction PQ, the angle BPQ being 




Fig. 34. 

equal to ANM. It is also possible to place the prism so that a 
ray, when refracted will meet the opposite face at such an angle 
that the ray will be either absorbed in, or reflected by, the surface 
(Art. 150). 

LENSES 
152. A lens is a refractive medium bounded by two surfaces 
through which light passes; one of the surfaces is spherical (that 
is, a part of a spherical surface) and the other is either a plane or a 



SB 




Fig. 35. 



part of a spherical surface. Lenses are classified according to the 
nature of these surfaces, there being six different kinds, shown in 
section in Fig. 35 and numbered from 1 to 6. Before naming 



106 



ELEMENTS OF PHYSICS 



§4 



them, refer to (7) , which shows two curves P and Q and a right 
hne MN that may be considered to be the trace of a plane, as the 
surface of a flat table top. The curve P may be regarded as 
the inside surface of a saucer; it is said to be concave with respect 
to the plane (or Hne) MN; the curve Q may be regarded as 
the inside surface of a saucer that has been turned upside down, 
and it is said to be convex with respect to the plane (or line) 
MN. Bearing these definitions in mind, lens (1) is a double 
convex lens, because, if laid on a flat surface with side B down, 
the upper side A will be convex; and if laid with side A down, 
the upper side B will be convex. Lens (2) is called plano-convex, 
because, if laid on a plane surface with side B down, the upper 
surface (which is flat) will be a plane, and if laid with side A 
down, the upper surface B will be convex. Lens (3) is crescent- 
shaped and is called a meniscus; if laid with side B down, the 
upper surface A will be concave, and if turned upside down (rest- 
ing on the points a and 6), the upper surface B will be convex. 
Lens (4) is called double-concave, because, whichever side is 
down, the other will be concave. Lens (5) is called plano- 
concave, side A being flat and B concave to it. Lens (6) is called 




Fig. 36. 



concavo-convex, because, if side B is down, side A will be con- 
cave, and if side A is down, side B will be convex. Strictly 
speaking, lens (3) might also be called concavo-convex, but on 
account of its crescent shape, it is called a meniscus, and the term 
concavo-convex is restricted to the form shown in (6) . 

Referring to Fig. 36, in both (a) and (6), C and C are the cen- 
ters of the spherical surfaces, CA being the radius of the surface 
A and C'B the radius of the surface B. The line CC joining the 
two centers is called the principal axis of the lens; in Fig. 35, XY 
is the principal axis. The two surfaces A and B are called the 
faces of the lens. Every lens has a point 0, Fig. 36, (a) and (&), 
called the optic center, which is situated between the faces in the 



§4 LIGHT 107 

case of the double convex or double concave lenses. If the radii 
CA and C'B are equal, the optic center is midway between the 
faces; in any case, it lies on the principal axis. 

A lens may be considered as composed of an infinite number of 
prisms; the explanations in Art. 151 therefore cover the principle 
of the lens. In general, the rays transmitted by any convex lens 
are caused to converge, while those transmitted by a concave lens 
are caused to diverge. The effects of a double convex lens (which 
is the most common form) and of a double concave lens on a 
beam of rays parallel to the principal axis are shown in (a) and 
(b), Fig. 36. In (a), if the lens is properly ground, the refracted 
rays meet in a point F, which is called the principal focus. In (b) 
the emergent rays cannot meet; but if their lines of direction be 
produced backwards, they will meet in a point F, as shown, which 
is the principal focus; the emergent rays appear to come from this 
point. The distance FO between the principal focus and the 
optic center is called the focal length. The distance ah is called 
the diameter of the lens. The thinner the lens for the same 
diameter the flatter will be its faces, the longer will be the radii CA 
and C'B, and the longer will be the focal length. 

The rays of light from the sun are parallel. If a convex lens be 
held in front of a piece of paper in such a manner that the light 
rays from the sun are parallel to the principal axis of the lens and 
the distance between the paper and the lens is increased or de- 
creased until the rays converge in a point, the point is the princi- 
pal focus, and the rays are said to be focused. If the paper and 
lens are then both kept stationary, the paper will very soon get 
hot at the point where the rays are focused and a little later will 
take fire. A lens acting in this manner is called a burning glass. 
If a luminant be placed at the principal focus, the rays from it 
will emerge parallel to the principal axis. On the other hand, 
if the light be placed beyond the principal focus, the emergent 
rays will converge and form an image; this can be easily verified 
by holding a lens between the flame of a candle and a sheet of 
paper, moving the lens and the paper until they are so adjusted 
with respect to the flame that a distinct image of the flame can be 
seen on the paper. This is the principle of the camera, in which 
the lens is moved to get the image distinct, i.e., to get the object 
''in focus." The eye is a camera; but, in this case, instead of the 
lens moving, it automatically becomes thinner or thicker, ac- 
cording as the object is far or near from the eye. 



108 



ELEMENTS OF PHYSICS 



§4 



153. The Magnifying Lens or Simple Microscope. — Referring 
to Fig. 37, L is a double convex lens andF is its principal focus. 
An object AB is between the lens and F, and the rays enter the 
eye on the other side of the lens. The effect is as though the 
rays came directly from a magnified image at A'B'. Such a 
lens is a simple microscope or magnifying glass. Note that the 
apparent positions of A and A' and of B and B' are determined 
by extending the ray AO, which passes through the optic center 
of the lens, and tracing the ray BD which is parallel to the 



.-.-A^ 




4^. 



Fig. 37. 



principal axis and consequently a limiting ray, until it meets AO 
produced at G. Similarly, the limiting ray AC and the 
undeviated ray BO are traced until they meet at E. If, now, the 
emergent ray HE and the ray AOG are produced backwards, 
they will meet at A', and IG and BOE produced backwards 
will meet at B'. All other points in AB produce images between 
A' and B', and this complete image of the points in AB is what the 
eye sees. It will be noticed that if the emerging rays HE and IG 
are produced, they will meet at the principal focus F', since AC 
and BD are parallel to the principal axis. The lens must be so 
held that the points E and G will fall on the lens of the eye. 
The compound microscope is an arrangement of several lenses 
that greatly magnifies the object. It is used for measuring the 
length and thickness of fibers, counting bacteria in milk, etc. 
The limit of magnification for the simple microscope is about 



§4 



LIGHT 



109 



100 diameters; the limit for the compound microscope has 
not as yet been reached, but a power of 1200 diameters is not 
uncommon, and some instruments have a power of from 2500 to 
3000 diameters. 



DISPERSION OF LIGHT 

154. The Solar Spectrum. — Suppose that ah is a narrow sHt in 
an opaque screen R, Fig. 38 and that a thin beam of sun hght 
passes through the sht, impinges on the glass prism P along cd, is 
refracted to ef, from whence it emerges. The light rays from ah 
to cd and from cd to ef are parallel; but after leaving the prism 
at ef, they form a pencil of rays egjihf having the shape of a wedge. 




Fig. 38. 

They will intersect the screen /S in a rectangle gjih and exhibit all 
the colors of the rainbow. The colored outline thus formed 
on the screen S is called the solar spectrum; the colors are called 
the colors of the spectrum or the prismatic colors. 

155. An examination of the spectrum shows seven prominent 
colors, viz.: red, orange, yellow, green, blue, indigo, and violet; 
these are called the seven prismatic colors. Moreover, they are 
always arranged in the above order, with the red nearest the 
line ef, where the rays emerge from the prism, and the violet 
farthest away. The first letters of the words designating each of 
the seven colors are given on the margin of the spectrum in Fig. 
37, and show the position of colors in the spectrum; reading 
them downwards, it will be noticed that they form the word 
vihgyor, which will assist in remembering the colors and their 
order. It must not be inferred that these are the only colors or 



110 ELEMENTS OF PHYSICS §4 

that they are sharply differentiated; as a matter of fact, the 
spectrum shows every color, shade, hue, and tint, all blending 
into one another, but the colors mentioned are the most promi- 
nent, and their order is always as here given. 

The cut shows that the rays entering the red part of the spectrum 
are not as refrangible as those entering the regions above the red, 
the violet rays being much more refrangible than the red rays. 
As the result of experiment and careful measurement (by methods 
that cannot be explained here), it has been found that the wave 
lengths of the rays differ in accordance with the color produced, 
the wave length of the violet rays being the shortest and of the red 
rays the longest. Since the speed, or velocity, of all rays is the 
same, it follows that the longer the wave the smaller will be the 
number of vibrations per second in the ether; and the shorter the 
wave the greater will be the number of vibrations per second. 
Hence, the red rays not only have a greater wave length than the 
violet rays, but they also make a smaller number of vibrations 
per second. Color is therefore due to the wave length; and if 
the wave length is greater than that of the red ray or is shorter 
than that of the violet ray, the ray will make no impression on the 
eye, that is, it cannot be seen. It is likewise evident that each 
color has its own index of refraction. 

156. White Light and the Primary Colors. — Since all colors 
except white and black are included in the spectrum, it is now 
clear that white light as received from the sun is a compound or 
mixture of all the colors. Black is not a color at all — it indicates 
the absence of color. The process of resolving light into its 
component colors is called dispersion or decomposition of light. 

None of the colors of the spectrum can be decomposed into 
any other color or combination of colors; this can be shown by 
cutting a slit anywhere in the screen covered by the spectrum, as 
mn in the blue region. Fig. 38. Allowing the rays to pass from 
this slit through the prism P' to the screen S/ both the incident 
and emergent rays will be parallel, and the color on the screen S' 
will be exactly the same as that of the slit on S. From this it 
will be seen that each color has its own index of refraction. 

If the screen S be replaced by a series of plane mirrors so arranged 
that they will reflect the various rays received from the prism P 
to a common point, called the focus, the light at the focus will 
be white, thus proving that a combination of all the colors of the 
spectrum produces white. 



§4 LIGHT 111 

157. Visible and Invisible Rays. — Those rays which produce 
the sensation of color, those whose wave lengths are not longer 
than the red rays or shorter than the violet rays, are called the 
visible rays. It can be easily proved that white light contains 
other rays, some of which extend beyond the red and others 
beyond the violet; in other words, some of these rays are less 
refrangible than the red rays and others are more refrangible 
than the violet rays. These rays are called the invisible rays. 
The invisible rays beyond the red are frequently called the 
ultra-red rays, those beyond the violet the ultra-violet rays, 
ultra meaning beyond. 

The ultra-red rays are heat rays, and their presence may 
be proved by placing a thermometer in that part of the spectrum 
beyond the red, when a rise in temperature will be observed. 
The ultra-violet rays are actinic or chemical rays, and their 
presence is revealed by their action on the photographic plate. 

158. Primary Colors. — Of the seven principal colors of the 
spectrum, three — red, yellow, and blue — are called the primary 
colors, because by properly mixing these colors, any other color, 
hue, shade, or tint may be obtained, including white. 

159. Secondary and Tertiary Colors. — When two primary 
colors are mixed, the result is a secondary color; thus, red and 
yellow produce orange, yellow and blue produce green, and red 
and blue produce violet. The secondaries vary in shade and 
tint in accordance with the proportions of the primaries used. 

When two secondary colors are mixed, the result is called a 
tertiary color. A tertiary color is said to be a combination of 
four colors, because the two secondaries must have one color in 
common, which is counted as two colors, and these added to the 
two primaries make four colors; thus, green and orange produce 
brown = yellow + blue + yellow 4- red. Hence, brown may 
have about twice as much yellow as it has of red or blue, according 
to the proportions used. 

160. Natural Colors. — The color of an object as seen in a clear 
white hght is called its natural color. The reason that an object 
has color is because it absorbs all the visible hght rays except 
those that are necessary to produce the natural color, the latter 
being reflected to the eye. This fact is easily proved: Place a 
red rose in the path of the beam of hght between the slit ah and 
the prism P in Fig. 38; the rose will be red. If held against the 



112 ELEMENTS OF PHYSICS §4 

screen in the red part of the spectrum, it will still be red; but if 
placed in the yellow part or any part above the yellow, it will 
appear black, showing that it has absorbed the light rays, none of 
them being reflected. In other words, a red object reflects only 
red rays, a yellow object only yellow rays, and a blue object only 
blue rays. A green object reflects both yellow and blue rays, 
and these combine to produce green, etc. 

An object never appears in its natural color or colors unless 
viewed in a clear white light or in a light having the exact color 
of the object. It is for this reason, that it is practically impossible 
to match colors, shades, etc. under artificial light, because no 
artificial light has yet been produced that is a pure white. A 
white lily appears white in a white light, because it reflects all the 
component colors of white light; but if it be placed in a red 
light, it will appear to be red, and if placed in a blue light, it 
will appear to be blue, etc., reflecting in each case the rays that 
make up the light. 

The color of a substance, as a sample of paper, can be re- 
presented numerically by means. of the tint photometer, which 
measures the percentage of red, yellow and blue in light re- 
flected from the surface. This instrument is described in the 
Section on "paper testing". 

161. The Spectroscope; Spectrum Analysis. — Any substance 
— solid, liquid, or gaseous — that can be burned, giving off a gaseous 
flame, has a spectrum peculiar to the substance and characteris- 
tic of it. These spectrums have bright lines, the number of 
which and their color and position identify the substances. 
The spectrum of burning sodium shows, in addition to other lines, 
two conspicuous, bright yellow lines that are so close together as 
to appear almost as one line; consequently, whenever these two 
lines appear in a spectrum of a flaming substance, and in their 
proper position, it is certain that sodium is present in the substance. 
The flame of carbon shows two distinct lines, one green and the 
other indigo. Other substances have their own special character- 
istics, which have been examined and tabulated, and when burned 
may be identified by their spectrums. This method of identify- 
ing substance is called spectrum analysis; by means of it, the 
composition of the sun and the stars has been ascertained. 

162. The spectroscope is an instrument used in determining 
the color and position of the bright lines that are peculiar to the 



§4 



LIGHT 



113 



substance being analyzed. It consists of, see (a), Fig. 39, a 
prism P enclosed in a case from which light is excluded and to 
which are attached three tubes M, N, and T, light from the 




burning substance enters tube M through a very narrow slit s, 
passes through the lens I', which makes the rays parallel, then 
passes through the prism P, which disperses the rays, resolves 
them into their spectral colors, and refracts them through the 



114 ELEMENTS OF PHYSICS §4 

tube T to the eye. Before reaching the eye, they pass through the 
lens I", which converges them to a focus, beyond which they are 
magnified by the eyepiece, the tube T being a telescope. The 
tube N contains a scale at r, the light rays from which pass 
through the lens V", which makes the rays parallel; they strike 
the prism P, and are reflected through the telescope tube T to the 
same relative position as the magnified image of the spectrum 
received from the rays passing through the slit s, thus making 
it possible to measure with a high degree of precision the position 
of any bright lines that may occur in the spectrum of the sub- 
stance being analyzed. A perspective view is shown at (6), 
Fig. 39. As may be supposed, there are many makes of spectro- 
scopes, the simpler ones having no scale for measuring the posi- 
tions of the bright lines. 

By means of spectrum analysis, several of the chemical ele- 
ments have been discovered, notably helium, which was found in 
the sun and named before it was discovered as one of the con- 
stituents of the earth. In some cases, spectrum analysis can be 
relied on when chemical analysis fails; thus, human blood and the 
blood of a pig cannot be distinguished chemically, but they can be 
identified separately by the spectroscope. 

163. Complementary Colors. — If two colors when mixed pro- 
duce white they are said to be complementary. For instance, 
if a line be drawn across the spectrum of white light, and one of 
the two colors contains all the colors on one side of the line and the 
other color contains all the colors on the other side of the line, 
the two colors when mixed will produce white and are called 
complementary. 

The following interesting experiment will show the comple- 
mentarj^ colors: Place on a black surface a small square or disk 
of some bright color, as red, yellow, or blue, and gaze at it steadily 
for about a minute ; then look at a white wall or white surface of 
any kind, as a sheet of paper, and the outline of the surface 
looked at will immediately appear, but its color will be the com- 
plement of that of the object. If the object is red, the comple- 
ment on the white surface will be bluish-green, if yellow, it will be 
blue, and if green, it will be crimson, etc. The following are 
some of the colors and their complements: 

Red Orange Yellow Violet Green 

Bluish-green Greenish-blue Blue Greenish-yellow Crimson 



§4 LIGHT 115 

The explanation of this phenomenon is that the nerves of the 
eye which respond to the color of the object gazed at become 
fatigued and do not respond to the white, which includes all colors, 
until a short time after the gaze has been removed from the object; 
the remaining nerves, which respond instantly to the other colors, 
blend these colors and produce the complementary color. As 
will be naturally expected, the experiment will be more successful 
if the object is placed on a black surface, since the only rays re- 
flected to the eye will then be those from the object. 

164. Lengths and Vibrations of Light Waves.— As was previ- 
ously mentioned, the wave length and, consequently, the number 
of vibrations per second varies in accordance with the color. It is 
the rate of vibration that determines any particular color. The 
wave lengths have been measured for different colors, and know- 
ing the wave length for any color, the rate of vibration for that 
color may be found by dividing the velocity of light by the length 
of the wave. Thus, the wave length of a red ray 

Wave length in Vibration per 

inches seconds 

Red 0000268 441,000,000,000,000 = 441 X lO^^ 

Orange 0000248 476,000,000,000,000 = 476 X lO^^ 

Yellow 0000228 518,000,000,000,000 = 518 X lO^^^ 

Green 0000204 579,000,000,000,000 = 579 X lO^^ 

Blue .0000182 649,000,000,000,000 = 649 X lO^^ 

Indigo 0000175 675,000,000,000 000 = 675 X lO'^ 

Violet... 0000166 712,000,000,000,000 = 712 X 10'^ 

is .0000268 inch; the velocity of light is 186,400 miles = 186400 
X 63360 inches per second; hence, the number of vibrations per 
second required to make light visible as a red ray is 

~j||360^^^j„„„ 000,000 

which may be written 441 X 10^^ a number almost inconceiv- 
ably large. The wave lengths and the vibrations per second for 
the seven principal colors are given in the foregoing table. 

165. Mixing Pigments. — Any coloring material used for paint- 
ing or printing is called a pigment. Coloring materials used in 
the paper industry are usually dye-stuffs, which, when fixed on the 
fibei- have the properties of pigments. Mixing pigments to get a 
certain desired color is quite a different problem from that of 
mixing or blending colored lights to get the same color. For 
instance, when yellow hght is added to a particular shade of blue. 



116 ELEMENTS OF PHYSICS §4 

the result is white hght, because, according to Art. 163, the two 
colors are complementary. If a yellow pigment be added to a 
blue one, the resulting color will be green; the reason for this is 
that the yellow pigment absorbs the blue and violet, the blue 
pigment then absorbs the red and yellow, with the result that only 
the green is left to be reflected. The final result wUl be obtained 
regardless of which color is applied first; simply allow it to dry 
and then apply the other color on top of the first Exact re- 
productions of the spectral colors cannot be obtained with pig- 
ments, because the pigments themselves do not have exactly the 
same shades as the colors of the spectrum, called the spectral 
colors. 

166. Three-color Process. — What is called the three-color 
process in printing is a more or less successful attempt to reproduce 
objects in their natural colors when using only three colors of ink. 
When the area of a disk is divided into three sectors, and one 
sector is colored red, the second green, and the third blue-violet, 
and the disk is caused to revolve rapidly, the three colors will 
blend into one single color. The jiature of the color obtained will 
depend upon the relative areas of the sectors, and by varying 
these, any desired color may be obtained. Calling the three 
colors just mentioned the primary colors, the primary pigments 
are the complements of these, and are (in order) peacock blue, 
crimson, and light yellow. It is to be observed that when the 
three primary colors mentioned above are mixed, the result will 
be white; but when the three primary pigments are mixed, the 
the result is hlack, because they absorb all the colors of white 
light. If black is mixed with white, the result is gray. 

In printing, the three primary pigments are applied to white 
paper in the following manner: Three different photographs of 
the object, which may be a natural object or a painting, are made, 
a gelatine screen, transparent, of the same color as one of the 
primary colors being placed in front of the camera lens, a 
different color for each photograph, and care is taken to have each 
photograph exactly the same in size. Then halftone blocks are 
made in the usual way from these photographs. The colored 
print is then made by printing on white paper from one of 
these halftone blocks, using an ink that is of the same color as the 
complement of the color of the screen that was used when the 
photograph was taken. After the ink has dried, the sheet is run 
through the press again, one of the other halftone blocks being 




Fig. 40 



§4 LIGHT 117 

used, and the ink being the complement in color of the color of the 
screen used in taking the photograph. The process is again 
repeated, using the third halftone block. To secure good results, 
it is necessary that perfect registration be obtained, that is, the 
second and third printings must be exactly superposed on the 
first. Fig. 40 shows the various steps in the printing, the little 
squares showing the color (or colors) of the ink used. A is the 
first printing, C the second, and E the third and final printing. 
It will be instructive to examine A, B, and D under a magnifying 
glass, which will bring out more clearly faint outlines that appear 
clearly in C and E. 

167. Paper is colored for two reasons ; first to produce an approxi- 
mate white, and second to produce a definite color, as green or 
pink. Paper pulp usually has a yellowish tint, reflecting a pre- 
ponderance of yellow rays. To correct this a proper mixture of 
red and blue coloring matter is added, so that the light reflected 
by it will be complementary to the yellow and so compensate or 
neutralize it. Other colors can likewise be compensated, but 
the mixing of pigments or dyestuffs always is a step toward 
darkness. 

Coloring paper to a definite shade is done by applying the prin- 
ciples explained in Art. 166. It makes no difference whether the 
coloring matter is a pigment that mixes with the fibres, or a 
dyestuff that stains them; the color of the paper is determined by 
the rays left to be reflected after part have been absorbed by the 
colored fibres. 



ELEMENTS OF PHYSICS 

(PART 2) 



EXAMINATION QUESTIONS 

(1) Steam is cut off in an engine cylinder at fths stroke and 
expands to the end of the stroke. Assuming that the fall in 
pressure follows Boyle's law for increase in volume (which is 
approximately true), what will be the pressure at the end of 
the stroke, if the pressure at cut off is 126 lb. per sq. in. gauge? 

Ans. 38+ lb. per sq. in. gauge. 

(2) If 4.6 cu. ft. of air at 96°F. expand at constant pressure 
until the temperature becomes 53°F., what will be the volume? 

Ans. 4.244 — cu. ft. 

(3) A vessel holding 1.58 cu. ft. is filled with air at a pressure 
of 14.65 lb. per sq. in. abs.; if the temperature of the air is 
70°F. and it is heated to 700°F., what will be its pressure? 

Ans. 32.08 lb. per sq. in. abs. 

(4) If 2.66 cu. ft. of air at 62°F. and a pressure of 1 atmosphere, 
is compressed to .52 cu. ft., what will be the tension when the 
air has a temperature of 112°F.? 

Ans. 67.71 lb. per sq. in. gauge. 

(5) Taking the specific gravity of nitrogen as .971, what is 
the weight of 875 cu. ft. when the tension is 20 lb. per sq. in. 
abs. and the temperature is 80^F.? Ans. 85.186 lb. 

(6) Referring to the last example, what is the weight of 1 cu. 
ft. of nitrogen at 32°F. and a tension of 1 atmosphere? 

Ans. .078517 lb. 

(7) Referring to Questions 5 and 6, if 3.8 cu. ft. of nitrogen at 
60°F. and a tension of 1 atmosphere is heated at constant volume 
to 1000°F., (a) how many B.t.u. must be expended? (6) how many 
foot-pounds of work are equivalent to this? (c) what will be the 
tension? [ (a) 46.7 B.t.u. 

Ans. (6) 36,333 ft.-lb. 

I (c) 41.3 — lb. per sq. in. abs. 
§4 119 



120 ELEMENTS OF PHYSICS §4 

(8) If 23 pounds of water at 45° are mixed with 18 pounds at 
190° and a piece of ice weighing 3 pounds is placed in the mix- 
ture, what will be the temperature after the ice has melted and 
the entire mixture has the same temperature, not considering 
the vessel holding it? Ans. 93.61°. 

(9) A platinum ball weighing 15.708 oz. is heated to a tem- 
perature of 2700°F. ; it is then placed in a wrought-iron vessel 
containing 2 lb. 5 oz. of water. If the temperature of the 
vessel and the water is 70°F., what will be the temperature of 
the mixture? Ans. 104.3°F. 

(10) Eef erring to the last question, suppose the temperature 
of the ball had not been known, but the temperature of the mix- 
ture had been found to be 104°F., what would be the temperature 
of the ball? Ans. 2677°F. 

(11) How many gram calories are equivalent to (a) 2.571 
B.t.u.? (6) to 1 B.t.u.? Ans. ( (a) 647.9 gram cal. 

I (6) 252 gram cal. 

(12) What is the temperature of saturated steam in a soda 
pulp digester when the pressure is 110 lb. per sq. in. gauge? 

Ans. 344.1°F. 

(13) What is the total heat of 1 pound of steam in a condenser 
when the vacuum gauge reads 11.5 in.? Ans. 1141.2 B.t.u. 

(14) What is the total heat of 7 lb. of steam when the pressure 
is 60 lb. per sq. in. gauge? Ans. 8270 B.t.u. 

(15) What is (a) a standard candle? (6) what is the candle- 
power of a lamp that, when placed 112 in. from a screen, illu- 
minates the screen with the same intensity as a standard candle 
at a distance of 24 in.? Ans. 27| c.p. 

(16) What is meant (a) by diffusion of light? (6) by refraction 
of light? (c) by reflection of light? (d) what is the relation be- 
tween the angle of incidence and the angle of reflection? 

(17) What are (a) the primary colors? (6) what is meant by 
secondary and tertiary colors? (c) what causes color? 

(18) What (a) is a pigment? (6) If three pigments having 
the colors of the three primary colors are mixed, what is the 
result? (c) three beams of light having the colors of the three 
primary colors are mixed, what is the result? (d) What is the 
cause of the difference in the two results? 



INDEX 



Note — The paging begins with 1 in each section, and each section has its number printed 
on the inside edge of the headline of each page. To find a reference, as "Abbreviations on 
drawings, §3, pl6," glance through the volume until §3 is found and then find page 16. 



Abbreviation in division, §2, p31 

in multiplication, §2, p30 
Abbreviations on drawings, §3, pi 6 
Absolute index (of refraction), §4, pl03 

pressure, definition of, §4, p53 

temperature, §4, p61 

zero of pressure, §4, p54 

zero of temperature, §4, p61 
Abstract number, §1, p2 
Acceleration, definition of, §4, p8 

positive and negative, §4, p9 

unit of, §4, p9 
Accuracy in calculation, §2, pp27-33 

in measurements, §2, p28 

in numerical operations, §2, p29 
Actinic rays, §4, p91 
Action, line of, §4, pl9 
Acute angle, §2, p46 
Addition, definition of, §1, pll 

of compound numbers, §1, pl20 

of decimals, §1, p68 

of fractions, rule for, §1, p55 

of mixed numbers, §1, p56 

of monomials, §2, p4 

of numbers, §1, ppl3-18 

of polynomials, §2, plO 

rule for, §1, pl5 

sign of, §1, pll 

table of, §1, pl2 
Adhesion, definition of, §4, plO 
Adjacent angles, §2, p45 
Aggregation, signs of, §1, p76; §2, pl5 
Air, definition of, §4, p45 
Altitude of cone, §2, pl21 

of cylinder, §2, pill 

of parallelogram, §2, p61 

of prism, §2, pi 02 

of pyramid, §2, pl04 

of triangle, §2, p56 
Amount, definition of, §1, plOl 

rule to find, §1, plOl 
Amplitude, definition of, §4, p91 
Analysis, spectrum, §4, pi 12 
Aneroid barometer, description of, §4, p52 
Angle, acute and obtuse, §2, p46 

critical, for emergent ray, §4, pl03 

definition of, §2, p45 

inscribed, §2, p71 

of deviation, §4, plOl 



Angle of incidence, §4, pp97, 101 
of reflection, §4, p97 
of refraction, §4, plOl 
re-entrant, §2, p66 
right, definition of, §1, ppllO, 112; 

§2, p46 

sides of, §2, p45 

vertex of, §2, p45 

Angles, adjacent, §2, p45 

circular measure of, §2, p83 
equality of, §2, p54 

measurement of, §2, p46 
Angular measure, table of, §1, pllO 
Apothem, definition of, §2, p67 
Arabic notation, §1, pp3, 5 
Arc, length of ciicular, §2, p85 

(of circle), definition of, §2, p70 

of curve, §2, p44 
Arc, definition of, §1, pll4 
Area, entire, definition of, §2, plOS 

of any plane figure, §2, p93 

of circle, to find, §2, p77 

of cylinder, §2, pplll, 112 

of ellipse, §2, p92 

of fillet, §2, p88 

of frustum of cone, §2, pl23 

of parallelogram, §2, p62 

of prism, §2, pl03 

of pyramid, §2, pl05 

of regular polygons, §2, p68 

of ring, §2, p82 

of sector and segment, §2, p87 

of sphere, §2, pl25 

of zone, §2, pl26 
Arithmetic, definition of, §1, p2 
Arithmetical mean, §1, pl25 
Application, point of, §4, pl4 
Atmosphere, definition of, §4, p45 

pressure of, §4, p46 

value of, §4, p46 
Atom, definition of, §4, p2 
Attraction, capillary, §4, p39 
Average of numbers, §1, pl25 
Avoirdupois weight, table of, §1, pl08 
Axes, major and minor, of ellipse, §2, p91 
Axis of cone, §2, pl20 

of pyramid, §2, pl05 

of sphere, §2, pi 24 

of symmetry, §2, pl37 

principal, of lens, §4, pl06 



121 



122 



INDEX 



Back view, §3, p7 
Barometer, definition of, §4, p51 
Barometers, mercurial and aneroid, §4, p52 
Base, definition of, §1, pp98, 100 

of parallelogram, §2, p62 

of spherical sector, §2, pl27 

of triangle, §2, p51 

rule to find, §1, pplOO, 101, 102 
Bases of a solid, §2, plOl 

of cylinder, §2, pi 10 

of spherical segment, §2, pl26 
Beam of light, rays, §4, p92 
Beaum6 scale for hydrometer, §4, p37 
Belt, to find length of, pp80, 81 
Binomial, definition of, §2, p9 
Blow, definition of, §4, pl9 
Body, definition of, §4, pi 
Boiling point, table of, §4, p86 
Bore or bored, definition of, §3, pi 6 
Bottom view, §3, p7 
Boyle's law, §4, p57 
Brightness of light, §4, p96 
British imperial gallon, §1, pl09 

thermal unit, definition of, §4, p67 
Brittleness, definition of, §4, p6 
Broken and dotted lines, use of, §3, pl2 

line, §2, p44 

lines, use of, §3, pl2 
Bunsen's method of measuring intensity of 

light, §4, p96 
Buoyancy, definition of, §4, p32 
Bushel, British, §1, pllO 

Winchester, §1, pl09 



Calculation, accuracy in, §2, pp27-33 

definition of, §1, p2 
Calorie, definition of, §4, p67 

gram or small, §4, p68 

kilogram or large, §4, p68 
Cancelation, §1, p37 
Candlepower, definition of, §4, p95 
Candle, standard, §4, p95 
Capacity, heat, §4, p68 
Capillarity, definition of, §4, p39 
Capillary attraction, §4, p39 

attraction, examples of, §4, p41 
Center of gravity, §2, pi 40 

of gravity, to find experimentally, §2 
pl41 

lines, §3, pplO, 13 

of ellipse, §2, p91 

of polygon, geometrical, §2, p66 

of sphere, §2, pi 24 

of symmetry, §2, ppl38, 140 

optic, §4, pl06 
Centigrade degrees to Fahrenheit, §4, p44 

scale, §4, p43 
Central angle, §2, p71 



C. G. S. system, §4, pl2 
Chain discount, §1, pl03 
Charles' law, §4, p60 
Check for division, §1, p32 

for multiplication, §1, p26 
Chord, definition of, §2, p70 

of arc, to find, §2, p75 

of half the arc, to find, §2, p76 
Cipher, naught, or zero, §1, p5 
Circle, definition of, §2, p68 

to describe a, §2, p69 

to find area and circumference, §2, p77 

to find diameter or radius, §2, p77 
Circles, concentric and eccentric, §2, p81 

great and small, §2, pl25 

properties of, §2, pp71-74 
Circular arc, definition of, §2, p70 

arc, length of, §2, p85 

measure of angles, §2, p83 
Circumference, definition of, §2, p70 

of circle, to find, §2, p77 • 

of ellipse, §2, p92 
Circumscribed polygon, §2, p89 
Clearing equation of fractions, §2, p20 
Coefficient, definition of, §2, p3 

of conductivity, definition of, §4, p76 

of conductivity, table of, §4, p76 

of expansion, §4, p72 
Coefficients of expansion, table of, §4, p73 
Cohesion, definition of, §4, p9 
Collecting terms, definition of, §2, pl9 
Coloring paper, reason for, §4, pll8 
Color, natural, §4, pill 
Colors, complementary, §4, pi 14 

primary, §4, pill 

prismatic, §4, pi 09 

secondary and tertiary, §4, pill 
Combined pressures on submerged surface, 

§4, p29 
Common denominator, §1, p52 

divisor, greatest, §1, p43 

multiple, least, §1, p45 
Compasses, definition of, §2, p69 
Complementary colors, §4, pll4 
Complete divisor, §1, p93 
Complex fraction, §1, p62 
Composite number, definition of, §1, p35 
Compound fraction, §1, p64 

proportion, §1, p83 

microscope, §4, pl08 

number, definition of, §1, pl06 

numbers, addition of, §1, pl20 

numbers, division of, §1, pl23 

numbers, multiplication of, §1,' pl22 

numbers, reduction of, §1, ppl 16-120 

numbers, subtraction of, §1, pl21 
Compressibility, definition of, §4, p4 
Computation, definition of, §1, p2 
Concave lens, §4, pl06 

polygon, §2, p66 
Concavo-convex lens, §4, pl06 



INDEX 



123 



Concentric circles, §2, p81 

Concrete number, §1, p2 

Conditions, standard, definition of, §4, p36 

Conduction of heat, §4, p75 

Conductivity, coefficient of, table of, §4, 

p76 
Conductors of heat, good and poor, §4, p75 
Cone, area and volume of, §2, pl21 

definition of, §2, pl20 

frustum of, area and volume of, §2, 
pl23 

of light rays, §4, p92 

of revolution, §2, pl20 

right, §2, pl20 

slant height and altitude of, §2, pl21 

vertex and axis of, §2, pi 20 
Conical surface, §2, pl20 
Constants, definition of, §2, p32 
Constant velocity, §4, p8 
Convection, definition of, §4, p77 
Conventional sections, §3, p22 
Convex area of cylinder, §2, pill 
lens, §4, pl06 

polygon, §2, p66 
Cord, definition of, §1, pl08 
Cored, definition of, §3, pi 6 
Critical angle for emergent ray, §4, pl03 
Cross hatching, §3, pl7 

section, §2, pl02; §3, 23 
Crossed belt, to find length of, §2, p81 
Crown of pulley, §3, p23 
Cylinder, altitude of, §2, pill 

bases of, §2, pi 10 

convex area of, §2, pill 

definition of, §2, ppllO, 117 

development of, §2. pill 

diameter of, §2, pill 

of revolution, §2, pi 10 

projection of, §3, p25 

right, §2, pi 10 

surface of, §2, pi 10 

volume of, §2, pi 14 
Cylindrical ring, volume of, §2, pl42 
Cube root of numbers, §2, p33-39 
Cubes and fifth powers, table of, §2, p35 
Cubical contents, definition of, §2, pl03 
Cubic centimeter, §1, pi 15 

foot, definition of, §1, pill 

measure, table of, §1, pl08 
Curve or curved line, §2, p44 



D 



d., dia., or diam., definition of, §3, pl6 
Decimal fractions, §1, p67 

point, definition of, §1, p8 
Decimals, addition and subtraction of, §1, 
p68 

definition of, §1, pp8, 67 

division of, rule for, §1, p73 

multiphcation of, §1, p69 



Decimals, reading, §1, p9 

reduction of, to common fractions, §1, 

pp67, 68 
to reduce to fractions having a given 
denominator, §1, p75 
Decomposition of light, §4, pi 10 
Degree of equation, §2, p23 
Degrees (of arc), definition of, §1, pllO 
on hydrometer scale, §4, p37 
on thermometer scale, §4, p43 
Denominate number, definition of, §1, pl06 
Denominator, common and least common, 
§1, p52 
of a fraction, §1, p48 
reduction of fraction to a given, §1, 
p51 
Density and specific gravity, difference 
between, §4, p35 
definition of, §4, pl4 
Development of cylinder, §2, pill 
Deviation, angle of, §4, plOl 
Dew point, §4, p87 
Diagonal of parallelogram, §2, p61 

of square, length of, §2, p63 
Diameter of circle, definition of, §2, p70 
of circle, to find, §2, p77 
of cylinder, §2, pill 
of sphere, §2, pl24 
Diameters of ellipse, §2, p92 
Difference, definition of, §1. pl8 

(in percentage) definition of, §1, plOl 
rule to find, §1, pl02 
Diffusion of light, §4, pi 00 
Digits, §1, p5 
Dimension lines, §3, pl3 
Dimensions, over-all, §3, pl4 
Direct ratio, §1, p78 
Direction of motion, §4, p7 
Directrix, definition of, §2, pll7 
Discount, chain, §1. pl03 
definition of, §1, pl02 
rate, rule for equivalent, §1, pl03 
Dispersion of hght, §4, pi 10 
Dissolve, definition of, §4, p2 
Dividend, definition of, §1, p28 

trial, §1, p92 
Divisibility, definition of, §4, p2 

of numbers, §1, p36 
Divisible, definition of, §1, p35 
Division, abbreviation in, §2, p31 
by a power of 10, §1, p67 
check for, §1, p32 
definition of, §1. p27 
long, §1, p31 

of compound numbers, §1, pl23 
of decimals, rule for, §1, p73 
of fractions, rule for, §1, p62 
of monomials, §2, p8 
of polynomials, §2, ppl3-15 
rule for, §1, p33 
short, §1, p29 



124 



INDEX 



Division, signs of, §1, p28 
Divisor, complete, §1, p93 

definition of, §1, p28 

greatest common, §1, p43 

trial, §1, p92 
Dollar, definition of, §1, pll2 
Dotted lines, use of, §3, pl2 
Double-concave lens, §4, pl06 
Double sign, meaning of, §2, p24 
Dozen, definition of, §1, pi 13 
Drawing, definition of, §3, p2 

perspective, §3, p2 

projection, §3, p2 

scale of, §3, pl5 

scaling a, §3, pl6 

to scale, §3, pl4 

working, §3, p8 
Drawings, abbreviations on, §3, pl6 

notes on, §3, pl6 

reading, directions for, §3, pp25-28 
Dry measure, table of, §1, pl09 

vapor, §4, p83 
Ductility, definition of, §4, p6 

E 

Eccentric circles, §2, p81 
Edges of a solid, §2, plOl 
Elasticity, definition of, §4, p4 
Element, definition of, §2, ppllO, 120 
Elevation, front, §3, p6 

rear, §3, p7 

side, §3, p6 
Ellipse, definition of, §2, p91 

diameters of, §2, p92 

foci of, §2, p91 

periphery and area of, §2, p92 

vertexes of and center of, §2, p91 
Energy, definition of, §4, pl6 

potential and kinetic, §4, pl6 

total, §4, pl7 
Entire area, definition of, §2, pl03 
Equality sign, definition of, §1, pll 
Equation, clearing, of fractions, §2, p20 

definition of, §2, pl8 

degree of, §2, p23 

members of, §2, pl8 

quadratic, formula for roots of, §2, p25 

roots of, §2, p25 

transposing terms of, §2, pl9 
Equations, identical and independent, §2, 
pl8 

linear, §2, p24 

of first degree, §2, p24 

of second degree, §2, p24 

quadratic, §2, p23 

quadratic, solution of, §2, p25 

transformation of, §2, pl9 
Equilibrium, definition of, §4, pl8 
Equivalent discount rate, rule for, §1, pl03 
Ether, definition of, §4, p78 



Even number, definition of, §1, p35 
Expansibility, definition of, §4, p4 
Expansion, coefiicient of, §4, p72 

linear, surface and cubic, §4, p72 
of bodies by heat, §4, p72 
Exponent, definition of, §1, p65 
Exponents, literal and numerical, §2, p4 
Extension, definition of, §4, pi 



f. or fin., definition of, §3, pl6 
Faced, definition of, §3, pl7 
Faces of lens, §4, pl06 

of solid, §2, plOl 
Factors, definition of, §1, p22 

prime, §1, p36 
Fahrenheit degrees, to centigrade, §4, p44 

scale, §4, p43 
Fifth powers and cubes, table of, §2, p35 

root of numbers, §2, p33-39 
Figure, order of, §1, p6 
Figures, definition of, §1, p5 

significant, §2, p27 

significant, number required, §2, p32 

similar, §2, pl31 

symmetrical, §2, pl37 
Fillet, area of, §2, p88 
fin. or finish, tool, definition of, §3, pl7 
First degree equations, §2, p24 
Fits, shrinking, §4, p74 
Fixed, definition of, §4, p6 
Flexibility, definition of, §4, p6 
Float, why some bodies, §4, p32 
Fluid, definition of, §4, p2 

ounce, §1, pl09 
Foaming, cause of, §4, p88 
Foci of ellipse, §2, p91 
Focal length of lens, §4, pl07 
Focus, principal, of lens, §4, pl07 
Focussed, meaning of, §4, pl07 
Foot, cubic, definition of, §1, pill 

of perpendicular, §2, p55 

square, definition of, §1, pill 
Force, definition of, §4, p9 

measure of, §4, plO 

various names of, §4, p9 
Formula, definition of, §2, pi 

prismoidal, §2, pl08 
Formulas, application of, §2, p22 
Fraction and integer, product of, §1, p68 

common, to reduce a, to a decimal, §1, 
p74 

complex, §1, p62 

to simplify a, §1, p63 

compound, §1, p64 

denominator of, §1, p48 

mixed, §1, p75 

numerator of, §1, p48 

power of, §1, p65 

proper or improper, §1, p49 



INDEX 



125 



Fraction, reduction of integer to improper, 
§1, p52 
reduction of mixed number to improper, 

§1, p52 
reduction of, to a given denominator, 

§1, p51 
signs of, §2, pl6 
terms of, §1, p49 
to divide by a fraction, §1, p61 
to divide by an integer, §1, p60 
value of, §1, p49 
Fractions, addition of, rule for, §1, p55 
common or vulgar, §1, p48 
decimal, §1, p67 
how to write, §1, p50 
multiplication of, §1, p58 
reduction of, §1, p50 
reduction of, to common denominator, 

§1, p52 
reduction of, to lowest terms, §1, p51 
subtraction of, rule for, §1, p57 
sum of two, §1, p55 
Front elevation, §3, p6 

view, §3, p6 
Froth, cause of and remedies for, §4, p89 
Frustum of cone, §2, pl22 

of pyramid, volume of, §2, pl06 
Full lines, use of, §3, pll 
Fusion, latent heat of, §4, p81 

G 

g, value of, §4, pl3 

Gain or loss per cent., rule for, §1, pl04 

Gallon, United States, wine, and imperial, 

§1, pl09 
Gas, definition of, §4, p2 

permanent, definition of, §4, p44 
Gases, mixture of, §4, p64 

perfect and imperfect, §4, p44 
specific gravity of, §4, p35 
Gauge pressure, §4, p54 
Gauges, vacuum, §4, p47 
Gay-Lussac's law, §4, p60 
G. C. D., definition of, §1, p43 
Generatrix, definition of,' §2, pi 17 
Geometrical center of polygon, §2, p66 
Glass, magnifying, §4, pl08 
Gram, definition of, §1, pll5 
Gravity, center of, §2, pl40 

specific, definition of, §4, p33 
specific, formulas for, §4, pp34, 35 
Great circle, §2, pl25 
Greatest common divisor, §1, p43 
Great gross, definition of, §1, pll3 
Grind, definition of, §3, pl7 
Gross, definition of, §1, pUS 



H 



' Hardness, definition of, §4, p6 
Heat capacity, §4, p68 



Heat capacity, conduction of, §4, p75 

definition of, §4, p66 

dynamical theory of, §4, p78 

expansion of bodies by, §4, p72 

latent and sensible, §4, p81 

mechanical equivalent of, §4, p67 

quantity of, §4, p67 

radiant, §4, p78 

specific, §4, p68 

transmission or propagation, §4, p75 
Hectare, definition of, §1, pll4 
Height of arc, to find, §2, p75 

of half the arc, to find, §2, p76 
Homologous, definition of, §2, pl32 

lines, properties of, pl33-137 
Horizontal line, §2, p46; §4, plO 
Humidity, relative, §4, p87 
Hydrometer, description of, §4, p36 

use of, §4, p38 
Hydrostatic machine, §4, p23 
Hydrostatics, definition of, §4, pl8 

I 

Identical equations, §2, pl8 

Image or picture, §3, pi 

Imaginary quantity, definition of, §2, p25 

Impenetrability, definition of, §4, p4 

Imperial gallon, British, §1, pl09 

Imperfect powers, §1, p90 

Improper fraction, §1, p49 

fraction, reduction of integer to, §1, p52 
fraction, reduction of mixed number to, 
§1, p52 
Impulse or blow, definition of, §4, pl8 
Incidence, angle of, §4, pp97, 101 
Incident ray, §4, plOl 
Independent equations, §2, pl8 
Indestructibility, definition of, §4, p5 
Index, absolute, of refraction, §4, pl03 
of refraction, §4, pl03 
of root, §1, p89 
Inertia, definition of, §4, p5 
Inferior characters, §2, p22 
Inscribed angle, §2, p72 

polygon, §2, p89 
Insulators (of heat), §4, p75 
Integer and fraction, product of, §1, p58 
definition of, §1, p8 
to divide by a fraction, §1, p61 
reduction of, to improper fraction, §1, 
p62 
Integral, definition of, §1, p9 
Intensity of light, §4, p95 
Intercept, definition of, §2, p71 
Intercepted arc, §2, p7l 
Interior angles of polygon, §2, p49 
Intersect, definition of, §2, p45 
Invisible rays, §4, pill 
Inverted plan, §3, p7 
Inverse ratio, §1, p78 
Involution, §1, p64 



126 



INDEX 



Kilo or kilogram, definition of, §1, pll5 
Kinetic energy, definition of, §4, pi 6 



Latent heat, definition of, §4, p81 

heat of fusion, §4, p81 

heat of vaporization, §4, p81 
Lateral sides of solid, §2, plOl 
Law, Boyle's or Mariotte's, §4, p57 

Boyle's and Gay-Lussac's combined, 
§4, p63 

Gay-Lussac's or Charles', §4, p60 

general, of all machines, §4, p24 

Pascal's, §4, p21 
Laws governing radiation, §4, p79 
Least common denominator, §1, p52 

common multiple, §1, p45 
L.G.M., definition of, §1, p45 
Length of circular arc, §2, p85 
Lens, definition of, §4, pl04 

faces of, §4, pl06 

focal length of, §4, pl07 

kind of, §4, pl06 

principal focus, of, §4, pl07 
Letters, use of, to designate lines, §2, p44 
Light, brightness and intensity of, §4, p95 

diffusion of, §4, plOO 

dispersion or decomposition of, §4, 
pllO 

emission of, §4, p91 

rays, §3, pi; §4, p91 

rays always straight, §4, p92 

rays, beam, cone, or pencil of, §4, p92 
Light, nature of, §4, p90 

reflection from curved surfaces, §4, 
p99 

reflection from plane surfaces, §4, 
p97 

refraction of, §4, plOl 

velocity of, §4, p93 

visible and invisible rays of, §4, pill 

wave length of, §4, p91 

waves, lengths and vibrations of, §4, 
pll5 

white, §4, pi 10 
Like numbers, §1, p2 

terms, §2, p5 
Liquid, definition of, §4, p2 

measure, table of, §1, pl09 
Linear equations, §2, p24 

measure, table of, §1, pl07 
Line, broken, §2, p44 

curved, §2, p44 

horizontal, vertical, and plumb, §2, 
p46 

mathematical, §2, p43 

right, §2, p44 

straight, definition of, §2, p43 



Line, vortical, horizontal or plumb, §4, plO 

of action, definition of, §4, pl9 
Lines, broken, use of, §3, pl2 

broken and dotted, use of, §3, pl2 

center, §3, pplO, 13 

designation of, by letters, §2, p44 

dimension, §3, pl3 

dotted, use of, §3, pl2 

full, use of, §3, pll 

parallel, §2, p45 

perpendicular, §2, p45 

section, §3, pl9 

thickness of, §3, pll 

weight of, §3, pll 
Liter, definition of, §1, pi 15 
Long division, §1, p31 
Longitudinal section, §2, pl02 
Luminous bodies, §4, p91 



M 



Machines, general law of all, §4, p24 
Machine, hydrostatic, §4, p23 
Magnifying glass, §4, pl08 
Major and minor axes of ellipse, §2, p91 
Malleability, definition of, §4, p6 
Mariotte's law, §4, p57 
Mass, definition of, §4, pll 
Mathematics, definition of, §1, p2 
Mathematical formula, definition of, §2, pi 

line, definition of, §2, p43 
Matter, definition of, §4, pi 

general properties of, §4, pi 

specific properties of, §4, p6 

three forms of, §4, pi 
Mean, arithmetical, §1, pl25 
Measure, angular, table of, §1, pllO 

circular, §2, p83 

cubic, table of, §1, pl08 

dry, table of, §1, pl09 

liquid, table of, §1, pl09 

linear, table of, §1, pl07 

square, table of, §1, pl07 

time, table of, §1, pll3 
Measurements, accuracy in, §2, p28 

units of, §1, pllO 
Measure of force, §4, plO 
Mechanical equivalent of heat, §4, p67 
Medium (for light), definition of, §4, p92 
Melting point, table of, §4, p86 
Members of an equation, §2, pl8 
Meniscus (lens), §4, pl06 

definition of, §4, p41 
Mensuration, definition of, §2, p43 
Mercurial barometer, description of, §4, 

p62 
Mercury, weight of 1 cu. in., §4, p46 
Meter, definition of, §1, pi 13 
Metric system, §1, ppll3-116 

ton, §1, pll5 
Microscope, simple and compound, §4, plOS 



INDEX 



127 



Minuend, definition of, §1, pl8 
Minus, definition of, §1, pl8 
Minutes (of arc), definition of, §1, pllO 
Mixed fractions, §1, p75 
number, §1, p49 
numbers, how to write, §1, p50 
numbers, addition of, §1, p56 
number, definition of, §1, p9 
numbers, division of, §1, p62 
numbers, multiplication of, §1, p59 
number, reduction of, to improper 

fraction, §1, p52 
number, to divide by a mixed number, 

§1. p62 
number, to divide by an integer, §1, 
p61 
Mixtures, temperature of, §4, p70 
Mixture of gases, §4, p64 
Mobility, definition of, §4, p5 
Molecule, definition of, §4, p2 
Money, United States, table of, §1, pll2 
Monomial, definition of, §2, p9 
Motion, definition of, §4, p7 

direction of, §4, p7 
Multiple, least common, §1, p45 
prime, §1, p45 
of a number, §1, p34 
Multiplicand, definition of, §1, p22 
Multiplication, abbreviation in, §2, p30 
check for, §1, p26 
definition of, §1, p22 
rule for, §1, p26 
sign of, §1, p22 
by a power of 10, §1, p66 
of compound numbers, §1, pl22 
of decimals, §1, p69 
of fractions, rule for, §1, p59 
of mixed numbers, §1, p59 
of monomials, §2, p7 
of polynomials, §2, ppll-13 
of numbers, §1, pp24-26 
table, §1, p23 
Multiplier, definition of, §1, p22 

N 

Names of periods, §1, p7 
Natural color, §4, pill 
Naught, cipher, or zero, §1, p5 
Negative and positive quantities, §2, p2 
Normal, definition of, §4, p21 

pressure on submerged surface, §4, p22 
Non-luminous bodies, §4, p91 
Notation, Arabic, §1, pp3, 5 

definition of, §1, p3 

Roman, §1, pp3, 4 
Notes on drawings, §3, pl6 
Number, abstract, §1, p2 

concrete, §1, p2 

definition of, §1, pi 

dividing into periods, §1, p6 



Number, mixed, §1, p49 

multiple of, §1, p34 

reading a, §1, p7 

root of a, §1, p89 

significant part of, §2, p27 

value represented by figures of a, §1, p6 

composite, definition of, §1, p35 

denominate, definition of, §1, pl06 

even, definition of, §1, p35 

mixed, definition of, §1, p9 

odd, definition of, §1, p35 

prime, definition of a, §1, p35 

simple and compound, definition of, 
§1, pl06 

whole, definition of, §1, p8 

mixed, i eduction of, to improper frac- 
tion, §1, p52 

mixed, to divide by a mixed number, 
§1, p62 

mixed, to divide by an integer, §1, p61 

of significant figures required, §2, p32 
Numbers, 

addition of, §1, ppl3-18 
average of, §1, pl25 
cube and fifth root of, §2, p33 
divisibility of, §1, p36 
like, §1, p2 
division of, §1, p27 
multiplication of, §1, pp24-26 
names of, §1, pp3, 4 
powers of, §1, p64 
properties of, §1, pp34— 38 
subtraction of, §1, pl8 
unlike, §1, p2 

compound, addition of, §1, pl20 
compound, division of, §1, pl23 
compound, multiplication of, §1, pl22 
compound, reduction of, §1, ppll6-120 
compound, subtraction of, §1, pl21 
mixed, addition of, §1, p56 
mixed, division of, §1, p62 
mixed, multiplication of, §1, p59 
mixed, subtraction of, §1, p57 
Numeration, definition of, §1, p3 
Numerator of a fraction, §1, p48 
Numerical operations, accuracy in, §2, p29 



O 



Obtuse angle, §2, p46 

Odd number, definition of, §1, p35 

Opaque body, §4, p92 

Open belt, to find length of, §2, p80 

Optic center, §4, 106 

Order of a figure, §1, p6 

of a figure, higher or lower, §1, p9 
Ordinate, definition of, §2, p94 
Orthographic projection, §3, p2 
Ounce, fluid, §1, pl09 
Over-all dimensions, §3, pl4 



128 



INDEX 



Paper, coloring, reason for, §4, pllS 
Parallel lines, §2, p45 

Parallelogram, altitude and diagonal of, §2, 
p61 

area and base of, §2, p62 

definition of, §2, p61 
Parallelograms, properties of. §2, p61 
Parallelopiped, definition of, §2, 102 
Partial vacuum, definition of, §4, p47 
Pascal's law, §4, p21 
Path of a body, §4, p7 
Pencil of light rays, §4, p92 
Penumbra, definition of, §4, p95 
Per cent, definition of, §1, p98 

gain or loss, rule for, §1, pl04 
Percentage, definition of, §1, p98 

rule to find, §1, 100 
Perch, definition of, §1, pl08 
Perfect gas, definition of, §4, p44 

powers, §1, p90 

vacuum, definition of, §4, p47 
Perimeter, definition of, §2, p58 

of polygon, §2, p48 

of triangle, §2, p58 
Periods, dividing a number into, §1, p6 

names of, §1, p7 
Periphery, definition of, §2, p70 
Permanent gas, definition of, §4, p44 
Perpendicular, foot of, §2, p55 

lines, §2, p45 
Picture or image, §3, pi 

plane, the, §3, pi 
Pigment, definition of, §4, pll5 
Pigments, mixing, §4, pll5 
Pipe, volume of, §2, pll5 
Piston, stroke of, §4, p49 
Photometers, description of, §4, pp95-97 
Plan, §3, p6 

inverted, §3, p7 
Plane, definition of, §2, p46 

the picture, §3, pi 

trace of, §3, p4 

of symmetry, §2, pl39 

figure, area of any, §2, p93 

figure, definition of, §2, p48 

figures, projection of, §3, p25 
Plano-convex lens, §4, pl06 
Plano-concave lens, §4, pl06 
Planed, definition of, §3, pi 7 
Pliability, definition of, §4, p6 
Plumb line, §2, p46; §4, plO 
Plus, definition of, §1, pll 

or minus sign, §2, p24 
Pneumatics, definition of, §4, p42 
Point, decimal, definition of, §1, p8 

dew, §4, p87 

of application, §4, pl4 

of intersection, §2, p44 

of sight, §3, p2 



Point of tangency, §2, p71 

projection of, §3, p25 
Pole, definition of, §2, pl24 
Polyedron, definition of, §2, pl02 
Polygon, convex, §2, p66 

definition of, §2, p4S 

geometrical center of, §2, p66 

inscribed and circumscribed, §2, p89 

interior angles of, §2, p49 

perimeter of, §2, p48 

re-entrant or concave; §2, p66 

regular, §2, p49 

sides and angles of, §2, p48 
Polynomial, arranging, according to de- 
scending powers, §2, p9 

definition of, §2, p9 
Polynomials, addition of, §2, plO 

division of, §2, ppl3-15 

multiplication of, §2, ppll-13 

subtraction of, §2, plO 
Polygons, area of regular, §2, p68 

names of, §2, p49 

properties of regulai, §2, p66 

table of regular, §2, p68 
Porosity, definition of, §4, p3 
Positive and negative quantities, §2, p2 
Potential energy, definition of, §4, pl6 
Power of a fraction, §1, p65 
Powers of 10, §1, p65 

of numbers, §1, p64 

perfect and imperfect, §1, p90 
Pressure, absolute, §4, p53 

absolute zero of, §4, p54 

at any depth, §4, p26 

definition of, §4, pl8 

depth required for a given, §4, p26 

due to weight of liquid, §4, p26 

from vacuum gauge reading, §4, p47 

gauge, §4, p54 

normal, on submerged surface, §4, p22 

of atmosphere, §4, p46 

on submerged surface due to weight of 
liquid, §4, p27 

specific and total, §4, pl9 
Pressures, combined, on submerged surface, 

§4, p29 
Primary colors, §4, pill 
Prime factors, §1, p36 

marks, §2, p21 

multiple, §1, p45 

or prime number, definition of a, §1, 
p35 
Priming, cause of and prevention of, §4, p90 
Principal axis of lens, §4, pl06 

focus of lens, §4, pl07 
Prism, altitude of, §2, pl02 

area and volume of, §2, pl03 

definition of, §2, pl02 

refraction through, §4, pl04 
Prismatic colors, §4, pl09 
Prismatoids and prismoids, §2, pl07 



INDEX 



129 



Prismoidal formula, §2, pl08 
Projection drawing, §3, p2 

of a cylinder, §3, p25 

of a point, §3, p26 

of plane figures, §3, p25 

orthographic, §3, p2 
Product, definition of, §1, p22 
Projection, definition of, §2, p58 

scenographic, §3, p2 
Projectors, §3, p4 
Propagation of heat, §4, p75 
Proper fraction, §1, p49 
Proportion, compound, §1, p83 

definition of, §1, p79 

direct and inverse, §1, p80 

how to read a, §1, p80 

law of, §1, p80 

names of terms of a, §1, p80 

three ways of writing, §1, p80 

to find value of unknown term, §1, p81 
Protractor, §2, p46 
Pull, definition of, §4, p9 
Pulley, crown of, §3, p23 
Pump, suction, description of, §4, p48 

work required to operate, §4, p49 
Push, definition of, §4, p9 
Pyramiid, area and volume of, §2, pl05 

axis of, §2, pi 05 

definition of, §2, pl04 

frustum of, volume of, §2, pl06 

regular, §2, pl05 

slant height of, §2, pl05 

vertex and altitude of, §2, pl04 



Quadrant, defintion of, §1, ppllO, 112 
Quadratic equation, formula for roots of, 
§2, p25 

equations, §2, p23, 24 

equations, solution of, §2, p25 
Quantity, definition of, §1, p2 

imaginary, definition of, §2, p25 

literal, §2, p3 

numerical, §2, p3 

numerical value of, §2, p3 

of heat, §4, p67 

unknown, §2, p23 
Quantities, positive and negative, §2, p2 
Quire, definition of, §1, pll3 
Quotient, definition of, §1, p28 

R 

R, value of for gas or air, §4, pp63, 64 
Radial section, §3, p23 

section, definition of, §2, pl43 
Radian, definition of, §2, p84 
Radiant, definition of, §4, p93 

heat, §4, p78 
Radiation, definition of, §4, p77 

laws governing, §4, p79 



Radius, definition of, §2, p70 
of arc, to find, §2, p75 
of circle, to find, §2, p77 
of sphere, §2, pl24 
Rate and rate per cent., difference between, 
§1, p99 
per cent., definition of, §1, p99 
rule for equivalent discount, §1, pl03 
rule to find, §1, plOO 
Ratio, definition of, §1, p78 

direct and inverse, §1, p78 
signs of, §1, p78 
terms of, §1, p78 
value of, §1, p78 
Ray, incident, §4, plOl 
of light, §4, p91 
refracted, §4, plOl 
Rays, actinic, §4, p91 

definition of, §2, pl31 

light, §3, pi 

of light, beam, cone or pencil of, §4, 

p92 
refrangible, §4, pl04 
ultra-red and ultra-violet, §4, pill 
visible and invisible, §4, pill 
Reading a number, §1, p7 
drawings, 

guard for movable jaw of vise, §3, p28 
mercer cell, the, §3, p30 
rotary drying furnace, §3, p33 
worm washer, §3, p33 
drawings, directions for, §3, pp25-28 
Ream, definition of, §1, pi 13 

or reamed, definition of, §3, pl6 
Rear elevation, §3, p7 

view, §3, p7 
Reciprocal, definition of, §2, p4 
Reduction ascending and descending, of 
fractions, §1, p51 
of common fraction to a decimal, §1, 

p74 
of compound numbers, §1, ppll6-120 
of decimals to common fractions, §1, 

pp67, 68 
of decimals to fractions having a given 

denominator, §1, p75 
of fractions, §1, p50 
of fractions to common denominator, 

§1, p52 
of fractions to lowest terms, §1, 

p51 
of integer to improper fraction, §1, p52 
of mixed number to improper fraction, 
§1, p52 
Re-entrant angle, §2, p66 

polygon, §2, p66 
Reflection, angle of, §4, p97 
of light, §4, pp97-100 
total, §4, pl03 
Refracted ray, §4, plOl 
Refraction, angle of, §4, plOl 



130 



INDEX 



Refraction, index of, §4, pl03 

of light, §4, plOl 

through prism, §4, pl04 
Refrangible rays, §4, pl04 
Regular polygon, definition of, §2, p49 

polygons, properties of, §2, p66 

polygons, table of, §2, p68 

pyramid, §2, pl05 
Relative, definition of, §4, p7 
Remainder, definition of, §1, pl8, 28 
Rest, definition of, §4, p6 
Rhomboid, definition of, §2, p61 
Rhombus, definition of, §2, p61 
Right angle, definition of, §1, ppllO, 112 
§2, 46 

cone, §2, pl20 

cylinder, §2, pi 10 

line, §2, p44 

parallelopiped, §2, pl02 

prism, §2, pl02 

section, §2, pl02 

triangle, properties of, §2, p53 
Rigidity, definition of, §4, p6 
Ring, area of, §2, p82 
Roman notation, §1, pp3, 4 
Root, index of, §1, p89 

of a number, §1, p89 

rule for square, §1, p95 
Roots of an equation, §2, p25 
Rules, trapezoidal and Simpson's, §2, p95 
Rumford's method of measuring intensity 
of light, §4, p96 



Satisfied, when equation is said to be, §2, 

pl9 
Saturated vapor, §4, p83 
Scale, Beaum6, §4, p37 

drawing, definition of, §3, pl5 

drawing to, §3, pl4 
Scales, centigrade and Fahrenheit, §4, p43 

definition of, §3, pl4 

special, §3, pl6 
Scaling a drawing, §3, pl6 
Scenographic projection, §3, p2 
Scraped, definition of, §3, pl7 
Secant, definition of, §2, p70 
Second degree equations, §2, p24 
Secondary colors, §4, pill 
Seconds (of arc), definition of, §1, pllO 
Section, cross, §3, p23 

lines, §3, pl9 

radial, §3, p23 

radial, definition of, §2, pl43 
Sectional view, §3, pl7 
Sections, conventional, §3, p22 

definition of, §3, pl7 

right, cross, and longitudinal, §2, pl02 

standard, §3, ppl9-21 

thin, §3, p21 



Sector, area of, §2, p87 

definition of, §2, p70 

spherical, base of, §2, pl27 
Segment, area of, §2, p87 

of circle, definition of, §2, p70 

of line, definition of, §2, p43 

spherical, §2, pl26 

spherical, volume of, §2, pl26 
Semicircle, definition of, §2, p70 
Semi-transparent body, §4, p92 
Sensible heat, definition of, §4, pSl 
Shadow, definition of, §4, p94 
Short division, §1, p29 
Shrinking fits, §4, p74 
Side elevation, §3, p6 

of square, length of, §2, p64 

view, §3, p6 
Sides of a solid, §2, plOl 

of angle, §2, p45 
Sight, point of, §3, p2 
Sign, double, meaning of, §2, p24 

of equality, definition of, §1, pll 

of multiplication, §1, p22 

of product of two monomials, §2, p7 

plus, definition of, §1, pll 

plus or minus, §2, p24 

of quotient, §2, p8 

of subtraction, §1, pl8 
Significant figures, §2, p27 

figures, number of, required, §2, p32 

part of a number, §2, p27 
Signs of a fraction, §2, pl6 

of aggregation, §1, p76; §2, pl5 

of division, §1, p28 

of ratio, §1, p78 
Similar figures, §2, pl31 

triangles, §2, p52 
Simple microscope, §4, pl08 

number, definition of, §1, pl06 
Simpson's rule, §2, p95 
Sink, why some bodies, §4, p32 
Siphon, description of, §4, p50 
Slant height of cone, §2, pl21 

height of pyramid, §2, pl05 
Small circle, §2, pl25 
Solar spectrum, §4, pl09 
Solid, definition of, §4, p2 

of revolution, volume of> §2, pl42 

sides, bases, and edges of, §2, plOl 
Specific gravity and density, distinction 
between, §4, p35 

gravity by hydrometer, §4, p37 

gravity, definition of, §4, p33 
formulas for, §4, pp34, 35 

gravity of gases, §4, p35 

heat, definition of, §4, p68 
table of, §4, pp69, 70 

pressure, definition of, §4, pl9 

weight, §4, pl4 
Spectroscope, description of, §4, pi 13 
Spectrum analysis, §4, pi 12 



INDEX 



131 



Spectrum, solar, §4, pl09 

Sphere, area and volume of, §2, pl25 

axis of, §2, pl24 

center of, §2, pl24 

definition of, §2, pl24 

radius and diameter of, §2, pl24 
Spherical sector, base of, §2, pl27 

segment, definition of, §2, pl26 

segment, volume of, §2, pl26 

surface, §2, pl24 
Square, diagonal of, length of, §2, p63 

foot, definition of, §1, pill 

measure, table of, §1, pl07 

root, rule for, §1, p95 

side of, length of, §2, p64 
Standard candle, §4, p95 

conditions, definition of, §4, p36 

sections, §3, ppl9-21 
Steam, dry, saturated, or superheated, §4, 
p83 

temperature of dry and saturated, §4, 
p83 

total heat of, §4, p84 
Straight line, definition of, §2, p43 
Stroke of piston or plunger, §4, p49 
Subscripts, §2, p22 
Subtended, definition of, §2, p71 
Subtraction, definition of, §1, pl8 

of compound numbers, §1, pl21 

of decimals, §1, p68 

of fractions, §1, p56 

of monomials, §2, p6 

of polynomials, §2, plO 

rule for, §1, pl9 

sign of, §1, pl8 
Subtrahend, definition of, §1, plS 
Suction, height of, §4, p49 

pump, description of, §4, p48 
Sum, definition of, §1, pll 

of two fractions, §1, p55 
Superheated vapor, §4, p83 
Superior characters, §2, p22 
Superposition, §2, p52 
Surface, conical, §2, pl20 

cylindrical, §2, pi 10 

definition of, §2, p47 

spherical, §2, pl24 

tension, §4, p88 
Symmetrical figures, §2, pl37 
Symmetry, axis of, §2, pl37 

center of, §2, pl38 

plane of §2, pl39 

with respect to a point, §2, pl40 



Table, addition, §1, pl2 
multiplication, §1, p23 
of coefficients of conductivity, §4, p76 
of coefficients of expansion, §4, p73 



Table of melting and boiling points, §4, 
p86 

of specific heats, §4, pp69, 70 
Tangency, point of, §2, p71 
Tangent, definition of, §2, p71 
Tap, definition of, §3, pl7 
Teaspoon, definition of, §1, pll5 
Temperature, absolute, §4, p61 

absolute zero of, §4, p61 

of mixtures, §4, p70 
Tenacity, definition of, §4, p6 
Tension of gases, definition of, §4, p42 

surface, §4, p88 
Terms, like and unlike, §2, p5 

of a fraction, §1, p49 

of a ratio, §1, p78 
Tertiary colors, §4, pill 
thds., definition of, §3, pl6 
Thermal conductivity, definition of, §4, p76 

unit, British, §4, p67 
Thermometer, definition of, §4, p42 
Thickness of lines, §3, pll 
Three-color process, §4, pll6 
Time, measures of, table of, §1, pll3 
Ton, long and short, difference between, §1, 

pl08 
Tonne or metric ton, §1, pll5 
Tool fin. or tool finish, definition of, §3, pl7 
Top view, §3, p6 
Torrecellian vacuum, §4, p46 
Torus, volume of, §2, pl42 
Total energy, §4, pl7 

pressure, definition of, §4, pl9 

reflection, §4, pl03 
Trace, front, §3, p4 

of a plane, §3, p4 

side, §3, p4 
Transformation of equations, §2, ppl9, 21 
Translucent body, §4, p92 
Transmission of heat, §4, p75 
Transparent body, §4, p92 
Transposing terms of equation, §2, pl9 
Trapezium, area of, §2, p65 

definition of, §2, p64 
Trapezoid, area of, §2, p64 

definition of, §2, p64 
Trapezoidal rule, §2, p95 
Trial dividend, §1, p92 

divisor, §1, p92 
Triangle, altitude of, §2, p56 

base of, §2, p51 

perimeter of, §2, p58 

relation between sides of, §2, p59 
Triangles, area of, §2, pp55-58 

classification of, §2, p50 

properties of, §2, p51 

proportionality of, §2, p52 

right, properties of, §2, p53 

similar, §2, p52 
Troy weight, §1, pl09 
Tube, volume of, §2, pi 15 



132 



INDEX 



u 



Ultra-red and ultra-violet rays, §4, pill 
Umbra, definition of, §4, p95 
Ungula, area of, §2, pi 13 

volume of, §2, pi 14 
Uniform velocity, §4, p8 
Unit, British thermal, §4, p67 

definition of, §1, pi 

of acceleration, §4, p9 

of work, §4, pl4 
United States gallon, §1, pl09 

money, table of, §1, pi 12 
Units of measurement, §1, pi 10 
Unknown quantity, definition of, §2, p23 
Unlike numbers, §1, p2 

terms, §2, p5 



Visible rays, §4, pill 

Visibility of objects, §4, plOl 

Visualizing objects from drawings, §3, p23 

Volume of any cylinder, §2, pi 14 
of cone, §2, pl21 

of cylindrical ring or torus, §2, pl42 
of frustum of cone, §2, pl23 
of pipe or tube, §2, pll5 
of prism, §2, pl03 
of pyramid, §2, pl05 
of sphere, §2, pl25 
of spherical sector, §2, pl27 
of spherical segment, §2, pl26 
of solid of revolution, §2, pl42 
of ungula, §2, pi 14 
of wedge, §2, pl09 



W 



Vacuum, definition of, §4, p33 

gauges, §4, pp47, 64 

perfect and partial, §4, p47 

Torricellian, §4, p46 

weight in, §4, p33 
Value of a fraction, §1, p49 

of ratio, §1, p78 
Vaporization, latent heat of, §4, p81 
Variable velocity, §4, p8 
Variables, definition of, §2, p32 
Velocity, definition of, §4, p7 

of light, §4, p93 

uniform or constant, §4, p8 

unit of, §4, p8 

variable, §4, p8 
Vertex of angle, §2, p45 

of cone, §2, pl20 

of pyramid, §2, pl04 
Vertexes of ellipse, §2, p91 
Vertical line, §2, p46 

line, definition of, §4, plO 
Vibgyor, meaning of, §4, pl09 
View, back or rear, §3, p7 

bottom, §3, p7 

front, §3, p6 

sectional, §3, pl7 

side, §3, p6 

top, §3, p6 
Views, names of, §3, p6 

number of, §3, p6 
Viscosity, definition of, §4, p88 



Water seeks its level, explanation of, §4, p28 

Wave length of light, §4, p91 

Waves, light, lengths and vibrations of, §4, 

pll5 
Wedge, volume of, §2, pl09 
Weight, avoirdupois, table of, §1, pl08 

definition of, §4, p5 

in vacuuna, §4, p33 

of air or gas, formula for, §4, p64 

of lines, §3, pll 

specific, §4, pl4 

troy, §1, pl09 

variation of, §4, pl3 
White light, §4, pi 10 
Whole number, definition of, §1, p8 
Winchester bushel, §1, pl09 
Wine gallon, §1, pl09 
Work, definition of, §4, pl5 

required to operate pump, §4, p49 

unit of, §4, pl4 
Working drawing, §3, p8 



Yard, definition of, §1, pi 10 
Year, length of, §1, pll3 



Zero, absolute, of pressure, §4, p64 
absolute, of temperature, §4, p61 
naught, or cipher, §1, p5 

Zone, area of, §2, pl26 



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